- (1) Given two random variables \(X\) and \(Y\) that are defined on the same probability space. The joint cumulative distribution function (CDF) \(F_{X,Y}\) is given by \[ F_{X,Y}(a,b) = P(\{X \le a, Y \le b\}) \quad \forall a,b \in \mathbb{R}. \]
- (2) The joint probability mass function (pmf) of two discrete random variables \(X\) and \(Y\) is \[ p_{X,Y}(x,y) = P(\{X = x, Y = y\}). \]
Hinglish anuvaad
Joint CDF aur joint PMF: X aur Y ek hi experiment ke do random variables hain. CDF poochta hai ki ek saath X≤a aur Y≤b ho. Discrete case mein joint PMF kisi ek exact pair (x,y) ka chance poochta hai.
📌 Formula: Joint CDF / joint PMF
\[F_{X,Y}(a,b)=P(X\le a,Y\le b),\qquad p_{X,Y}(x,y)=P(X=x,Y=y)\]English content ke steps
Step 1: CDF ka '≤' dono conditions par ek saath lagta hai. Isliye F(a,b) ek rectangle ke andar ki poori probability jodta hai.
Step 2: PMF ke (x,y) mein pehli coordinate X ki aur doosri Y ki value hai. P(X=x,Y=y) wahi outcomes count karta hai jo dono statements satisfy karein.
Step 3: Sab possible pairs ke joint PMF ko add karoge to 1 aayega. Isi joint table se aage marginal distributions nikalengi.
Toss a coin twice, and assume all outcomes are equally likely.
\(\Omega = \{(\omega_1, \omega_2) \mid \omega_1, \omega_2 \in \{H,T\}\}\), \(\omega = (\omega_1, \omega_2)\).
Define
Then,
\begin{align*} p_{X,Y}(0,0) &= P(\{X=0, Y=0\}) = \frac{1}{4} \\ p_{X,Y}(0,1) &= P(\{X=0, Y=1\}) = \frac{1}{4} \\ p_{X,Y}(1,0) &= P(\{X=1, Y=0\}) = \frac{1}{4} \\ p_{X,Y}(1,1) &= P(\{X=1, Y=1\}) = \frac{1}{4} \end{align*}Hinglish anuvaad
Do baar fair coin toss hota hai. X pehle toss par H ko 1 aur T ko 0 deta hai; Y doosre toss par H ko 1 aur T ko 0 deta hai. Chaar possible (X,Y) pairs ki probabilities dikhani hain.
📌 Formula: Joint PMF
\[p_{X,Y}(x,y)=P(X=x,Y=y)\]English content ke steps
Step 1: Sample space ke four ordered outcomes HH, HT, TH, TT hain. Har outcome ka chance 1/4 hai, kyunki source mein sab equally likely maana hai.
Step 2: Pair (0,0) ka matlab TT; (0,1) TH; (1,0) HT; (1,1) HH. Har pair ke liye exactly ek outcome hai.
Step 3: Isliye table mein chaaron entries 1/4 hain; 4×1/4=1 check bhi ho jata hai. Do tosses ki independence ko baad ke card mein alag se test karenge.
Roll a die: \(\Omega = \{1,2,\dots,6\}\).
Define
| \(\omega\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| \(X\) | 1 | 0 | 1 | 0 | 1 | 0 |
| \(Y\) | 0 | 1 | 1 | 0 | 1 | 0 |
(Note: The PDF has a slight variation in table values, corrected logically based on odd/prime definitions).
Hinglish anuvaad
Ek fair die ke result par X=1 jab face odd ho, warna 0; Y=1 jab face prime (2,3,5) ho, warna 0. Table se har (X,Y) pair ka chance nikaalna hai.
📌 Formula: Joint PMF ko outcomes gin kar nikaalo
\[p_{X,Y}(x,y)=\frac{\#\{\omega:X(\omega)=x,\ Y(\omega)=y\}}6\]English content ke steps
Step 1: Face 4,6 neither odd nor prime hain: (0,0) ke two outcomes, probability 2/6. Face 2 even aur prime hai: (0,1) ka 1/6.
Step 2: Face 1 odd lekin prime nahi: (1,0) ka 1/6. Faces 3,5 odd aur prime dono: (1,1) ka 2/6.
Step 3: Chaar entries ka sum (2+1+1+2)/6=1 hai. English note mein PDF se table correction ka remark hai; odd aur prime ki given definitions hi use ki gayi hain.
- (1) The distribution of \(X\) can be obtained from the joint distribution of \(X\) and \(Y\) as follows: \[ F_X(a) = P(\{X \le a\}) = P(\{X \le a, Y < \infty\}). \] Similarly, \(F_Y(b) = P(\{Y \le b\}) = P(\{X < \infty, Y \le b\})\).
- (2) The probability mass function of \(X\) is given by
\[
p_X(x) = P(\{X = x\}) = \sum_{y : p(x,y)>0} p_{X,Y}(x,y).
\]
Similarly, \(p_Y(y)\) is defined.
Example: Consider example 2. \[ p_X(0) = P(\{X=0\}) = P(\{X=0, Y=0\}) + P(\{X=0, Y=1\}) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}. \]
Hinglish anuvaad
Joint distribution se sirf X ki distribution chahiye ho, toh Y ki har possible value ko include karo; isi ko marginal kehte hain. Isi tarah sirf Y ke liye X ko include karte hain.
📌 Formula: Marginal PMF: doosre variable par sum
\[p_X(x)=\sum_y p_{X,Y}(x,y),\qquad p_Y(y)=\sum_x p_{X,Y}(x,y)\]English content ke steps
Step 1: CDF wale first remark mein Y<∞ likhne ka matlab finite real valued Y par koi restriction na rakhna hai; sirf X≤a ki condition bachi.
Step 2: Discrete PMF mein fixed x ki row ko Y ke sab y par add karo. Ye alag disjoint cases hain, isliye probabilities sum hoti hain.
Step 3: Die example mein X=0 ka matlab even: joint entries (0,0)=2/6 aur (0,1)=1/6. Dono jodne par p_X(0)=3/6=1/2.
Step 4: Agar Y ka marginal chahiye toh same table ka fixed y column add karna hai.
We say that \(X\) and \(Y\) are jointly continuous if \(F_{X,Y}(a,b)\) is continuous.
The function \(f_{X,Y}(x,y)\) is said to be the joint probability density function (pdf) of \(X\) and \(Y\) if
The marginal pdf of \(X\) & \(Y\) are defined as:
\[ f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy, \quad f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dx. \]Remarks: \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) dx dy = 1\).
- (1) If \(X\) and \(Y\) are jointly continuous, then \(X\) & \(Y\) are continuous R.V.s.
Proof: For \(a \in \mathbb{R}\), we have \begin{align*} F_X(a) &= P(\{X \le a\}) = P(\{X \le a, Y < \infty\}) \\ &= \int_{-\infty}^{a} \left( \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy \right) dx \\ &= \int_{-\infty}^{a} f_X(x) dx. \end{align*} Hence \(X\) is a continuous R.V. with pdf \(f_X(x)\).
Hinglish anuvaad
X,Y ke liye joint CDF aur joint density ki definitions di hain. Joint density f(x,y) ko region par double integrate karne se probability milti hai; Y par integrate karke f_X aur X par karke f_Y milta hai.
📌 Formula: Joint CDF / marginal density
\[F_{X,Y}(a,b)=\int_{-\infty}^{a}\int_{-\infty}^{b}f_{X,Y}(x,y)\,dy\,dx;\quad f_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\]English content ke steps
Step 1: F(a,b) mein integration limits x≤a aur y≤b ka rectangle banati hain. Full plane par density ka double integral 1 hota hai, kyunki kahin na kahin pair (X,Y) zaroor hoga.
Step 2: f_X ke liye x fixed rakho aur Y ki saari values ke over integrate karo. Fir f_X ko −∞ se a tak integrate karne par F_X(a) milta hai; ye English proof ke three lines hain.
Step 3: Isi tarah f_Y ke liye X ko integrate out karo. Ek nuance: sirf joint CDF continuous hona har distribution mein joint pdf ki guarantee nahi deta; yahan pdf wali formulas tab use karo jab joint density exist kare.
Suppose \(X\) and \(Y\) are jointly continuous random variables with density given by
\[ f_{X,Y}(x,y) = \begin{cases} c & \text{if } x^2 + y^2 \le R^2 \\ 0 & \text{otherwise} \end{cases} \]Find (1) \(c\) \quad (2) Find marginal density of functions of \(X\) and \(Y\).
Solution:
(1) We know \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X,Y}(x,y) dx dy = 1\).
(2)
\[ f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy = \frac{1}{\pi R^2} \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} dy = \frac{2\sqrt{R^2-x^2}}{\pi R^2} \quad \text{if } x^2 \le R^2. \]and \(f_X(x) = 0\) if \(x^2 > R^2\).
Hinglish anuvaad
Joint density disk x²+y²≤R² ke andar constant c aur bahar zero hai. c nikaalo aur phir X ki marginal density nikaalo. Circle aur vertical slice ki figure isi integration region ko dikhati hai.
📌 Formula: Normalization / marginal integration
\[c=\frac1{\pi R^2},\qquad f_X(x)=\frac{2\sqrt{R^2-x^2}}{\pi R^2}\quad(|x|\le R)\]English content ke steps
Step 1: Part (1): Total probability 1 honi chahiye. Circle ka area πR² hai aur density uspar c constant hai. Isliye cπR²=1, aur c=1/(πR²), assuming R>0.
Step 2: Part (2): Ek x fix karne par disk equation se y²≤R²−x² aata hai; y ki limits −√(R²−x²) se +√(R²−x²) hain.
Step 3: In limits par c ko dy se integrate karne ka matlab vertical slice ki length 2√(R²−x²) ko c se multiply karna hai. |x|>R par disk ki slice hi nahi, isliye density zero.
We say that two random variables \(X\) and \(Y\) are independent if
\[ F_{X,Y}(a,b) = F_X(a)F_Y(b) \quad \text{for all } a,b \in \mathbb{R}. \]Hinglish anuvaad
Random variables X aur Y independent tab kehlate hain jab har pair a,b ke liye joint CDF unke separate CDFs ke product ke barabar ho.
📌 Formula: Random variables ki independence
\[F_{X,Y}(a,b)=F_X(a)F_Y(b)\quad\text{for every }a,b\]English content ke steps
Step 1: Left side dono thresholds ko saath satisfy karne ka chance hai. Right side unke individual chances ka product hai.
Step 2: Condition har real a,b ke liye chahiye; ek particular pair par equality milna sufficient nahi. Agle remarks mein joint pdf/PMF se ise kaise test karein, woh diya hai.
- (1) Suppose \(X\) and \(Y\) are continuous with joint pdf \(f_{X,Y}(x,y)\). Then
\begin{align*}
\int_{-\infty}^{a} \int_{-\infty}^{b} f_{X,Y}(x,y) dx dy &= F_{X,Y}(a,b) = F_X(a)F_Y(b) \\
&= \int_{-\infty}^{a} f_X(x)dx \int_{-\infty}^{b} f_Y(y)dy \\
&= \int_{-\infty}^{a} \int_{-\infty}^{b} f_X(x)f_Y(y) dx dy.
\end{align*}
\(\implies f_{X,Y}(x,y) = f_X(x)f_Y(y)\).
Conversely, if this holds then \(F_{X,Y}(a,b) = F_X(a)F_Y(b) \implies X\) and \(Y\) are independent.
Where \(f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dy\) and \(f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) dx\). - (2) Similarly, \(X\) and \(Y\) are independent discrete R.V.s iff \[ p_{X,Y}(x,y) = P(\{X=x, Y=y\}) = p_X(x)p_Y(y). \]
Hinglish anuvaad
Continuous case mein independent X,Y ki joint density marginal densities ke product ke barabar hoti hai. Discrete case mein exactly isi test ko joint PMF aur marginal PMFs se karte hain.
📌 Formula: Continuous / discrete independence tests
\[f_{X,Y}(x,y)=f_X(x)f_Y(y),\qquad p_{X,Y}(x,y)=p_X(x)p_Y(y)\]English content ke steps
Step 1: Continuous proof mein independent joint CDF ko F_X(a)F_Y(b) likha. Dono CDFs ko density integrals se replace kiya; product ko double integral bana diya.
Step 2: Joint density ka integral aur product density ka integral har rectangular region par same ho, toh densities almost everywhere same hoti hain. Isliye pdf equality technical sense mein almost everywhere hoti hai.
Step 3: Conversely factorization ko integrate karne se CDF product milta hai. Discrete case mein har (x,y) entry p_X(x)p_Y(y) honi chahiye, tab independence milegi.
- (1) In example 1, \(X\) and \(Y\) are independent.
- (2) In example 2, \(X\) and \(Y\) are not independent.
As \(P(\{X=0, Y=0\}) = \frac{2}{6} = \frac{1}{3}\).
But \(P(\{X=0\}) = \frac{1}{2}\), \(P(\{Y=0\}) = \frac{1}{2} \implies P(\{X=0\})P(\{Y=0\}) = \frac{1}{4} \neq \frac{1}{3}\). - (3) In example 3, \(X\) and \(Y\) are not independent.
As \(f_{X,Y}(x,y) = \frac{1}{\pi R^2} \neq f_X(x)f_Y(y)\) for \(x^2+y^2 < R^2\).
Hinglish anuvaad
Teen examples pichhle cards ka independence rule check karte hain: two coin tosses independent, odd/prime die variables dependent, aur circular region mein chosen point ki X/Y coordinates dependent.
📌 Formula: Joint distribution ko marginals ke product se compare karo
\[p_{X,Y}(x,y)\overset{?}{=}p_X(x)p_Y(y)\quad\text{or}\quad f_{X,Y}(x,y)\overset{?}{=}f_X(x)f_Y(y)\]English content ke steps
Step 1: Example (1): Coins mein har joint entry 1/4 aur har marginal 1/2 hai. 1/4=(1/2)(1/2), sab pairs mein test pass hota hai.
Step 2: Example (2): Die par P(X=0,Y=0)=2/6=1/3, lekin P(X=0)P(Y=0)=(1/2)(1/2)=1/4. Ek mismatch hi independence reject karne ke liye enough hai.
Step 3: Example (3): Disk mein X fixed hone par allowed Y interval badalta hai. Joint constant density ka product of marginals se mismatch hai, isliye X,Y independent nahi.
Let \(X\) and \(Y\) be two independent standard normal random variables. Find \(P(\{X^2 + Y^2 \le 1\})\).
Solution:
We have
Since \(X\) and \(Y\) are independent, the joint probability density function (pdf) is given by
\[ f_{X,Y}(x,y) = \frac{1}{2\pi} e^{-\frac{x^2+y^2}{2}}, \quad -\infty < x, y < \infty. \]Therefore,
\begin{align*} P(\{X^2 + Y^2 \le 1\}) &= \iint_{x^2+y^2 \le 1} \frac{1}{2\pi} e^{-\frac{x^2+y^2}{2}} dx dy \\ &= \frac{1}{2\pi} \int_{\theta=0}^{2\pi} \int_{r=0}^{1} e^{-\frac{r^2}{2}} r dr d\theta \quad \text{(Polar coordinates: } x=r\cos\theta, y=r\sin\theta, dxdy=rdrd\theta) \\ &= \int_{0}^{1} e^{-\frac{r^2}{2}} r dr \\ &= \int_{0}^{1/2} e^{-t} dt \quad \text{(Let } t = \frac{r^2}{2} \implies dt = r dr) \\ &= -e^{-t} \Big|_{0}^{1/2} \\ &= 1 - e^{-1/2}. \end{align*}Hinglish anuvaad
X aur Y independent standard normal hain. Unit circle X²+Y²≤1 ke andar point (X,Y) aane ki probability nikaalni hai.
📌 Formula: Independence + polar change of variables
\[P(X^2+Y^2\le1)=\int_0^{2\pi}\int_0^1\frac{e^{-r^2/2}}{2\pi}r\,dr\,d\theta=1-e^{-1/2}\]English content ke steps
Step 1: Standard normal ka ek-variable density e^(−x²/2)/√(2π) hai. Independence se X aur Y ki densities multiply karke e^(-(x²+y²)/2)/(2π) milti hai.
Step 2: X²+Y²=r² aur unit disk mein r=0 se 1, angle θ=0 se 2π. Polar substitution x=r cosθ, y=r sinθ mein area element dxdy=r dr dθ hota hai; extra r bhoolna nahi.
Step 3: θ integral 2π factor deta hai jo denominator 2π se cancel hota hai. t=r²/2 put karo, dt=r dr; limits 0 aur 1/2. Integral ∫₀¹ᐟ² e^(−t)dt=1−e^(−1/2).
Let \(X_1\) and \(X_2\) be two random variables with joint pdf \(f_{X_1, X_2}(x_1, x_2)\). Suppose \(Y_1 = g_1(X_1, X_2)\) and \(Y_2 = g_2(X_1, X_2)\) such that
\[ X_1 = h_1(Y_1, Y_2) \quad \text{and} \quad X_2 = h_2(Y_1, Y_2). \]And
\[ J(x_1, x_2) = \det \begin{bmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{bmatrix} \quad \text{(Jacobian determinant matrix)}. \]Then the joint pdf of \((Y_1, Y_2)\) is given by
\[ f_{Y_1, Y_2}(y_1, y_2) = f_{X_1, X_2}(x_1, x_2) |J(x_1, x_2)|^{-1} \quad \text{(absolute value)} \]with \(x_1 = h_1(y_1, y_2)\), \(x_2 = h_2(y_1, y_2)\).
Hinglish anuvaad
Do variables (X₁,X₂) ko functions g₁,g₂ se naye variables (Y₁,Y₂) mein badal rahe hain. Inverse X₁=h₁(Y₁,Y₂), X₂=h₂(Y₁,Y₂) aur Jacobian se nayi joint density milti hai.
📌 Formula: Change of variables / Jacobian
\[f_{Y_1,Y_2}(y_1,y_2)=f_{X_1,X_2}(h_1(y),h_2(y))\left|\det\frac{\partial(y_1,y_2)}{\partial(x_1,x_2)}\right|^{-1}\]English content ke steps
Step 1: Pehle old-to-new mapping aur inverse solve karo. Jacobian matrix ki entries ∂yᵢ/∂xⱼ hain; determinant local area ka scaling factor hai.
Step 2: Probability conserve karne ke liye old density ko inverse Jacobian magnitude se multiply karo; determinant ka sign area ke liye matter nahi karta, isliye absolute value hai.
Step 3: Result mein old x₁,x₂ ko inverse h₁(y),h₂(y) se replace karna aur naye variables ki allowed range check karna zaroori hai. Formula one-to-one differentiable mapping aur nonzero Jacobian wali region par lagu hota hai.
Suppose \(X_1\) and \(X_2\) are two independent standard normal random variables. Find the joint probability density function of \((Y_1, Y_2)\) where \(Y_1 = X_1 + X_2\) and \(Y_2 = X_1 - X_2\). Also, find \(f_{Y_1}(y_1)\) and \(f_{Y_2}(y_2)\).
Solution:
Let \(Y_1 = X_1 + X_2\) and \(Y_2 = X_1 - X_2\).
\(|J|^{-1} = \frac{1}{2}\).
\begin{align*} f_{Y_1, Y_2}(y_1, y_2) &= f_{X_1, X_2}(x_1, x_2) |J(x_1, x_2)|^{-1} \\ &= f_{X_1, X_2}\left(\frac{y_1 + y_2}{2}, \frac{y_1 - y_2}{2}\right) \cdot \frac{1}{2} \\ &= \frac{1}{4\pi} e^{-\frac{1}{2} \left\{ \left(\frac{y_1+y_2}{2}\right)^2 + \left(\frac{y_1-y_2}{2}\right)^2 \right\}} \\ &= \frac{1}{4\pi} e^{-\frac{1}{2} \left\{ \frac{2(y_1^2 + y_2^2)}{4} \right\}} \\ &= \frac{1}{4\pi} e^{-\frac{y_1^2 + y_2^2}{4}} \\ &= \frac{1}{\sqrt{4\pi}} e^{-\frac{y_1^2}{4}} \cdot \frac{1}{\sqrt{4\pi}} e^{-\frac{y_2^2}{4}}. \end{align*}Since \(f_{Y_1, Y_2}(y_1, y_2) = f_{Y_1}(y_1) f_{Y_2}(y_2)\), we get:
\[ f_{Y_1}(y_1) = \frac{1}{\sqrt{4\pi}} e^{-\frac{y_1^2}{4}}, \quad f_{Y_2}(y_2) = \frac{1}{\sqrt{4\pi}} e^{-\frac{y_2^2}{4}}. \]\(\implies Y_1\) and \(Y_2\) are independent.
Hinglish anuvaad
X₁,X₂ independent standard normal hain. Y₁=X₁+X₂ aur Y₂=X₁−X₂ ke joint density aur dono marginals nikaal kar independence dekhni hai.
📌 Formula: Inverse transform + Jacobian + factorization
\[f_{Y_1,Y_2}(y_1,y_2)=\frac1{4\pi}e^{-(y_1^2+y_2^2)/4}=f_{Y_1}(y_1)f_{Y_2}(y_2)\]English content ke steps
Step 1: Dono transformation equations ko add/subtract karke X₁=(Y₁+Y₂)/2 aur X₂=(Y₁−Y₂)/2 milte hain.
Step 2: Forward Jacobian [[1,1],[1,−1]] ka determinant −2 hai. Isliye absolute inverse factor 1/2 hai; old joint normal density 1/(2π) se product 1/(4π) hota hai.
Step 3: Identity (y₁+y₂)²+(y₁−y₂)²=2(y₁²+y₂²) exponent ko simplify karti hai. Isse joint density do N(0,2) densities ke product mein split hoti hai.
Step 4: Har factor doosre variable par integrate karne se respective marginal milta hai. Joint = marginals ka product har y₁,y₂ par hai, so Y₁,Y₂ independent.
Let \(X_1, X_2\) be two random variables with joint distribution \(f_{X_1, X_2}(\cdot, \cdot)\). Suppose \(g : \mathbb{R}^2 \rightarrow \mathbb{R}\) is a function. Then
\[ E[g(X_1, X_2)] = \iint_{\mathbb{R}^2} g(x_1, x_2) f_{X_1, X_2}(x_1, x_2) dx_1 dx_2. \]Hinglish anuvaad
Joint density f_{X₁,X₂} aur kisi function g(x₁,x₂) ko diya hai. g(X₁,X₂) ki expectation nikaalne ke liye g ko joint density se weight karke poore plane par double integrate karte hain.
📌 Formula: Function of two variables ki expectation
\[E[g(X_1,X_2)]=\iint_{\mathbb R^2}g(x_1,x_2)f_{X_1,X_2}(x_1,x_2)\,dx_1\,dx_2\]English content ke steps
Step 1: Ek variable mein E[g(X)]=∫g(x)f_X(x)dx tha. Ab do variables hain, isliye har pair (x₁,x₂) ke value g(x₁,x₂) ko us pair ki density se weight karte hain.
Step 2: Integral poore allowed support par hai; outside support density zero hoti hai. Is formula ke liye expectation ka finite/defined hona bhi check karna hota hai.
- (1) If \(g(x_1, x_2) = h(x_1)\) then \begin{align*} E[h(X_1)] &= \iint_{\mathbb{R}^2} h(x_1) f_{X_1, X_2}(x_1, x_2) dx_1 dx_2 \\ &= \int_{-\infty}^{\infty} h(x_1) \left( \int_{-\infty}^{\infty} f_{X_1, X_2}(x_1, x_2) dx_2 \right) dx_1 \\ &= \int_{-\infty}^{\infty} h(x_1) f_{X_1}(x_1) dx_1. \end{align*}
- (2) \(E[X_1] = \int_{-\infty}^{\infty} x_1 f_{X_1}(x_1) dx_1\).
- (3) \(E[h(X_1) + k(X_2)] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (h(x_1) + k(x_2)) f_{X_1, X_2}(x_1, x_2) dx_1 dx_2\)
\begin{align*}
&= \iint h(x_1) f_{X_1, X_2}(x_1, x_2) dx_1 dx_2 + \iint k(x_2) f_{X_1, X_2}(x_1, x_2) dx_1 dx_2 \\
&= \int h(x_1) f_{X_1}(x_1) dx_1 + \int k(x_2) f_{X_2}(x_2) dx_2 \\
&= E[h(X_1)] + E[k(X_2)].
\end{align*}
In particular, \(E[X_1 + X_2] = E[X_1] + E[X_2]\).
Similarly, \(E[h_1(X_1, X_2) + h_2(X_1, X_2)] = E[h_1(X_1, X_2)] + E[h_2(X_1, X_2)]\) (Linearity of expectation). - (4) Suppose \(X_1\) and \(X_2\) are independent. Then \begin{align*} E[h(X_1) k(X_2)] &= \iint h(x_1) k(x_2) f_{X_1}(x_1) f_{X_2}(x_2) dx_1 dx_2 \\ &= \left[ \int h(x_1) f_{X_1}(x_1) dx_1 \right] \left[ \int k(x_2) f_{X_2}(x_2) dx_2 \right] \\ &= E[h(X_1)] E[k(X_2)]. \end{align*} In particular, \(E[X_1 X_2] = E[X_1] E[X_2]\) if \(X_1, X_2\) are independent.
- (5) If \(g(X_1, X_2) = e^{t_1 X_1 + t_2 X_2}\) for some \(t_1, t_2 \in \mathbb{R}\). \[ M_{X_1, X_2}(t_1, t_2) = E[e^{t_1 X_1 + t_2 X_2}] \quad \text{moment generating fn of } (X_1, X_2). \] If \(X_1\) and \(X_2\) are independent then \[ M_{X_1, X_2}(t_1, t_2) = M_{X_1}(t_1) M_{X_2}(t_2). \]
Hinglish anuvaad
Expectation of a function ke five remarks hain: marginal se one-variable expectation, mean, sums par linearity, independent variables ke products, aur joint MGF.
📌 Formula: Linearity / independence / joint MGF
\[E[h(X_1)+k(X_2)]=E[h(X_1)]+E[k(X_2)];\quad M_{X_1,X_2}(t_1,t_2)=E[e^{t_1X_1+t_2X_2}]\]English content ke steps
Step 1: (1) g(x₁,x₂)=h(x₁) ho to h ke andar x₂ nahi. x₂ par pehle joint density integrate karoge to f_{X₁}(x₁) marginal milta hai.
Step 2: (2) h(x₁)=x₁ choose karne se ordinary mean E[X₁]=∫x₁f_{X₁}(x₁)dx₁ milta hai.
Step 3: (3) Sum ke integrals separate karne se E[h(X₁)+k(X₂)]=E[h(X₁)]+E[k(X₂)]. Is linearity ke liye independence required nahi hai.
Step 4: (4) Product expectation tab factor hoti hai jab independence se joint density f_{X₁}f_{X₂} ho. Independence sufficient hai; particular functions ke product equality ka milna alone independence prove nahi karta.
Step 5: (5) Joint MGF e^(t₁X₁+t₂X₂) ki expectation hai. Independence mein exponential e^(t₁X₁)e^(t₂X₂) aur expectation factor hoti hai, toh individual MGFs ka product milta hai.
Let \(X_1, X_2, \dots, X_n\) be independent \(Ber(p)\) random variables and \(Y\) be \(Bin(n, p)\) random variable. Show that
\[ X_1 + X_2 + \dots + X_n \overset{d}{=} Y \][We say \(X\) has same distribution as \(Y\), denoted by \(X \overset{d}{=} Y\), if \(F_X(t) = F_Y(t)\) \(\forall t \in \mathbb{R}\).
FACT: If \(M_X(t) = M_Y(t)\) for some \(t \in \mathbb{R}\), then \(X \overset{d}{=} Y\)].
Solution:
Note \(X_i \sim Ber(p)\), that is,
\(E[e^{tX_i}] = p e^t + (1-p)\) for \(i=1,\dots,n\).
\begin{align*} \implies M_{X_1+\dots+X_n}(t) &= E[e^{t(X_1+\dots+X_n)}] \\ &= \prod_{i=1}^{n} E[e^{tX_i}] \quad \text{ (since independent)} \\ &= (p e^t + (1-p))^n \end{align*}On the other hand, \(Y \sim Bin(n, p)\). \(P(Y=k) = \binom{n}{k} p^k (1-p)^{n-k}\) for \(k=0,1,\dots,n\).
\begin{align*} M_Y(t) &= E[e^{tY}] = \sum_{k=0}^{n} e^{tk} P(Y=k) \\ &= \sum_{k=0}^{n} \binom{n}{k} (p e^t)^k (1-p)^{n-k} \\ &= (p e^t + (1-p))^n. \end{align*}Thus, \(M_{X_1+\dots+X_n}(t) = M_Y(t)\).
By the FACT, \(X_1 + \dots + X_n \overset{d}{=} Y\).
Hinglish anuvaad
n independent Ber(p) variables ke sum ka distribution Bin(n,p) random variable Y jaisa prove karna hai. English mein same distribution ka meaning CDF equality aur MGF method diya hai.
📌 Formula: Bernoulli sum / Binomial MGF
\[M_{X_1+\cdots+X_n}(t)=(1-p+pe^t)^n=M_{\operatorname{Bin}(n,p)}(t)\]English content ke steps
Step 1: Har Xᵢ=1 probability p aur 0 probability 1−p, so E[e^(tXᵢ)]=pe^t+1−p. Independent hone se sum ki MGF in n identical factors ka product hai.
Step 2: Y ke Binomial PMF ko ∑e^(tk)C(n,k)pᵏ(1−p)ⁿ⁻ᵏ mein substitute karo. Binomial theorem se wahi (pe^t+1−p)ⁿ milta hai.
Step 3: Isliye dono MGFs zero ke aas-paas ke interval par equal hain aur distributions same hain. English FACT ki correction: sirf ek t par M_X(t)=M_Y(t) hona sufficient nahi; especially t=0 par har existing MGF 1 hoti hai.
- (1) If \(X\) and \(Y\) are two independent random variables, then
\[
Var(X+Y) = Var(X) + Var(Y).
\]
Proof:
\begin{align*}
Var(X+Y) &= E[(X+Y)^2] - (E[X+Y])^2 \\
&= E[((X - E[X]) + (Y - E[Y]))^2] \\
&= E[(X - E[X])^2] + 2E[(X - E[X])(Y - E[Y])] + E[(Y - E[Y])^2] \\
&\overset{\text{ind}}{=} Var(X) + 2E[X - E[X]]E[Y - E[Y]] + Var(Y) \\
&= Var(X) + Var(Y).
\end{align*}
In general, \(Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y)\).
\(Cov(X,Y) = E[(X - E[X])(Y - E[Y])]\). - (2) If \(X_1 \sim \mathcal{N}(\mu_1, \sigma_1^2)\) and \(X_2 \sim \mathcal{N}(\mu_2, \sigma_2^2)\) and independent then \[ X_1 + X_2 \sim \mathcal{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2). \]
Hinglish anuvaad
Independent X,Y ke sum ki variance unki variances ka sum hai; general case mein extra 2Cov(X,Y) bhi hota hai. Independent normal variables ka sum phir normal hota hai.
📌 Formula: Variance addition / covariance
\[\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)\]English content ke steps
Step 1: Proof mein X+Y ke mean ko subtract karke [(X−E[X])+(Y−E[Y])]² expand kiya. Do squared terms individual variances hain aur beech ka cross term 2Cov(X,Y) hai.
Step 2: Independence par centered products ki expectation E[X−E[X]]E[Y−E[Y]]=0×0=0. Isliye covariance term vanish karti hai.
Step 3: Normal sum ke liye means μ₁+μ₂ aur independent variances σ₁²+σ₂² add hote hain. Normal-family result ko sirf variance addition se alone prove nahi kar sakte; distribution property bhi use hoti hai.
Let \(X \sim Bin(n,p)\). Find \(E[X]\) and \(Var(X)\).
Solution:
Let \(X_1, \dots, X_n\) be independent \(Ber(p)\) random variables.
Then we have
Hinglish anuvaad
X~Bin(n,p) ka mean aur variance nikaalna hai. Isko n independent Ber(p) indicators ke sum ke same distribution ke roop mein likha gaya hai.
📌 Formula: Binomial mean / variance
\[E[X]=np,\qquad\operatorname{Var}(X)=np(1-p)\]English content ke steps
Step 1: Pichhle problem se X aur X₁+⋯+Xₙ same distribution rakhte hain, so unka mean aur variance bhi same hai.
Step 2: Har Bernoulli mean 1·p+0·(1−p)=p; linearity se n terms ka mean np aata hai.
Step 3: Har Bernoulli ka E[Xᵢ²]=p kyunki 0²=0, 1²=1. Variance p−p²=p(1−p), aur independent sum par n variances add karke np(1−p).
H.W Problem: Let \(X_1, \dots, X_n\) be independent \(\exp(\lambda)\) and \(X \sim \text{Gamma}(n, \lambda)\).
Show that \(X \overset{d}{=} X_1 + \dots + X_n\) [Show that moment generating fns are same].
Find \(Var(X)\) and \(E[X]\).
Hinglish anuvaad
Homework mein n independent Exp(λ) waiting times ka sum Gamma(n,λ) ke same distribution ka dikhana hai; phir mean aur variance nikaalne hain.
📌 Formula: Exponential sum / Gamma MGF
\[M_{\sum_iX_i}(t)=\left(\frac\lambda{\lambda-t}\right)^n\ (t<\lambda);\quad E[\sum_iX_i]=\frac n\lambda;\quad\operatorname{Var}(\sum_iX_i)=\frac n{\lambda^2}\]English content ke steps
Step 1: Har Exp(λ) ka MGF λ/(λ−t) hai (rate λ>0). Independent sum ke liye in n MGFs ko multiply karo; result (λ/(λ−t))ⁿ hai.
Step 2: Isi rate convention ke Gamma(n,λ) MGF se equality zero ke aas-paas milti hai, isliye same distribution.
Step 3: Ek Exp(λ) ka mean 1/λ aur variance 1/λ². n independent copies ke mean aur variance ko add karke n/λ aur n/λ² milte hain.
- (1) If \(g(X) \ge 0\) for all \(X \in \mathbb{R}\) then \(E[g(X)] \ge 0\).
Follows from the definition of expectation. - (2) If \(g(X) \ge h(X)\) for all \(X \in \mathbb{R}\). Then \[ E[g(X)] \ge E[h(X)]. \] Follows from (1).
- (3) (Markov inequality) If \(X \ge 0\) then,
\[
P(\{X \ge a\}) \le \frac{E[X]}{a}.
\]
Proof:
Note \(\frac{X}{a} \ge I(\{X \ge a\})\).
\(\implies E\left[\frac{X}{a}\right] \ge E[I(\{X \ge a\})]\). \[ \implies P(\{X \ge a\}) \le \frac{E[X]}{a}. \] Note: \(I(\{X \ge a\}) = \begin{cases} 1 & \text{if } X \ge a \\ 0 & \text{otherwise} \end{cases}\)
\(E[I(\{X \ge a\})] = 1 \cdot P(X \ge a) + 0 \cdot P(X < a) = P(\{X \ge a\})\). - (4) (Chebyshev's inequality): Let \(X\) be a r.v. Then
\[
P(\{|X - E[X]| \ge a\}) \le \frac{Var(X)}{a^2}
\]
Proof:
We have \begin{align*} P(\{|X - E[X]| \ge a\}) &= P(\{|X - E[X]|^2 \ge a^2\}) \\ &\le \frac{E[(X - E[X])^2]}{a^2} \quad \text{(Using Markov inequality)} \\ &= \frac{Var(X)}{a^2}. \end{align*}
Hinglish anuvaad
Is card mein expectation ke order rules, nonnegative X par Markov bound, aur mean se deviation par Chebyshev bound with proofs hain.
📌 Formula: Markov / Chebyshev inequalities
\[P(X\ge a)\le\frac{E[X]}a\ (X\ge0,a>0);\qquad P(|X-E[X]|\ge a)\le\frac{\operatorname{Var}(X)}{a^2}\]English content ke steps
Step 1: (1) Agar g(X)≥0 har outcome par hai, to uska probability-weighted average E[g(X)] bhi ≥0. (2) g≥h mein g−h≥0 apply karke E[g]≥E[h] milta hai.
Step 2: (3) Markov proof mein event indicator I{X≥a} sirf 0 ya 1 hai. X≥a par X/a≥1, aur X<a par X/a≥0; isliye X/a≥I. Expectations lene par E[X]/a≥P(X≥a). a>0 required hai.
Step 3: (4) Chebyshev mein nonnegative variable Z=(X−E[X])² lo. |X−E[X]|≥a exactly Z≥a² hai. Z par Markov lagao: E[Z]/a²=Var(X)/a². Isliye Chebyshev Markov se aaya.
Suppose \(X\) denotes the number of Heads of tossing a fair coin 100 times. Estimate:
- (1) \(P(\{X > 80\})\)
- (2) \(P(\{40 \le X \le 60\})\).
Solution:
(1) (a) We use Markov inequality:
(b) Using Chebyshev's inequality:
\[ P(\{X > 80\}) \le P(\{|X - 50| \ge 30\}) \le \frac{Var(X)}{30^2} = \frac{25}{30 \times 30} = \frac{1}{36}. \](Here \(Var(X) = 100 \times p \times (1-p) = 25\)).
(2) We use Chebyshev's inequality:
\begin{align*} P(\{40 \le X \le 60\}) &= P(\{-10 \le X - 50 \le 10\}) \\ &= P(\{|X - 50| \le 10\}) \\ &= 1 - P(\{|X - E[X]| > 10\}) \\ &\ge 1 - \frac{Var(X)}{10^2} \\ &= 1 - \frac{25}{100} = \frac{3}{4}. \end{align*}Hinglish anuvaad
Fair coin 100 baar toss hota hai; X Heads ki count hai. X>80 ka upper bound aur 40≤X≤60 ka lower bound nikaalna hai.
📌 Formula: Markov and Chebyshev estimates
\[E[X]=50,\quad\operatorname{Var}(X)=25;\quad P(X>80)\le\frac1{36};\quad P(40\le X\le60)\ge\frac34\]English content ke steps
Step 1: X~Bin(100,1/2), so mean np=50 aur variance np(1−p)=25. Ye dono bounds mein input hain.
Step 2: (1a) X nonnegative hai; X>80 event, X≥80 ke andar hai. Markov se P(X>80)≤50/80=5/8. Ye upper bound hai, actual chance 5/8 nahi.
Step 3: (1b) X>80 par |X−50|≥30 zaroor hai. Chebyshev se P(X>80)≤25/900=1/36, jo tighter bound hai.
Step 4: (2) 40≤X≤60 ka matlab |X−50|≤10. Complement mein |X−50|>10 aata hai; Chebyshev bound ≤25/100=1/4, so target chance ≥3/4. Equality at endpoints ko source ke ≤ aur > notation ke according hi handle kiya hai.
Let \(X_1, \dots, X_n, \dots\) be independent and identically distributed random variables with finite mean \(\mu\) and variance \(\sigma^2\). Then, for \(\epsilon > 0\),
\[ P\left(\left\{ \left| \frac{X_1 + \dots + X_n}{n} - E[X_1] \right| > \epsilon \right\}\right) \rightarrow 0 \quad \text{as } n \rightarrow \infty. \]Proof:
For \(\epsilon > 0\), let \(S_n = \frac{X_1 + \dots + X_n}{n}\).
Then
Calculation for \(Var(S_n)\):
\begin{align*} Var(S_n) &= \frac{1}{n^2} Var(X_1 + \dots + X_n) \\ &\overset{\text{(indep.)}}{=} \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i) \\ &\overset{\text{(eq. dist.)}}{=} \frac{n \cdot Var(X_1)}{n^2} \\ &= \frac{\sigma^2}{n}. \end{align*}Hinglish anuvaad
Independent aur identically distributed Xᵢ ka finite mean μ aur variance σ² hai. Sample average true mean μ se ε se zyada door ho, uski probability n badhne par zero jaati hai.
📌 Formula: Chebyshev se Weak Law of Large Numbers
\[P(|S_n-\mu|>\epsilon)\le\frac{\sigma^2}{n\epsilon^2}\longrightarrow0,\qquad S_n=\frac1n\sum_{i=1}^{n}X_i\]English content ke steps
Step 1: Proof ka first step Sₙ=(X₁+⋯+Xₙ)/n define karna hai. Linearity aur identical means se E[Sₙ]=nμ/n=μ.
Step 2: Variance mein constant 1/n square hokar 1/n² hota hai. Independence se n variances add hoti hain, aur identical distribution se har variance σ²: Var(Sₙ)=nσ²/n²=σ²/n.
Step 3: Chebyshev bound Var(Sₙ)/ε²=σ²/(nε²) hai. ε fixed positive ho aur n→∞, to bound zero hota hai; probability bhi zero ki taraf jaati hai. Statement convergence in probability hai: finite n par sample average exactly μ hona required nahi.
- (1) Weak Law is also written as \(\frac{X_1 + \dots + X_n}{n} \overset{P}{\rightarrow} E[X_1]\) (Convergence in Probability).
- (2) SLLN (Strong law of large number): If \(X_1, X_2, \dots, X_n, \dots\) are i.i.d random variables with finite mean, then \[ \frac{X_1 + \dots + X_n}{n} \xrightarrow{a.s.} E[X_1] \quad \text{(almost sure sense)}. \] We skip the details here.
Hinglish anuvaad
Weak Law ko probability mein convergence Sₙ→μ likhte hain. Strong Law usi sample average ke μ par almost surely converge hone ka statement hai; source mein uska proof nahi diya.
📌 Formula: Weak versus Strong Law notation
\[S_n\xrightarrow{P}\mu\quad\text{(Weak)},\qquad S_n\xrightarrow{a.s.}\mu\quad\text{(Strong)}\]English content ke steps
Step 1: (1) Sₙ→ᴾμ ka meaning: har ε>0 par P(|Sₙ−μ|>ε)→0. Pichhle proof ne isi property ko Chebyshev se establish kiya.
Step 2: (2) Sₙ→ᵃ·ˢ·μ ka meaning: probability 1 wale outcome paths par averages ka sequence μ ki taraf jaata hai. Ye stronger conclusion hai; source mein i.i.d. finite mean assumption ke under statement diya hai, details skip ki hain.