We say a sequence of random variables \(\{X_n\}\) converges to \(X\) in distribution if
\[ P(X_n \le t) \longrightarrow P(X \le t) \quad \text{as } n \rightarrow \infty \]\(\forall t \in C(F_X)\) : the set of continuity pts of \(F_X\).
Notation: \(X_n \xrightarrow{d} X\) as \(n \rightarrow \infty\).
Hinglish anuvaad
Random variables Xₙ ka sequence X ko distribution mein converge karta hai jab unke CDFs Fₓₙ(t), X ke CDF Fₓ(t) ki taraf jaayen—un sab t par jahan Fₓ continuous hai.
📌 Formula: Convergence in distribution
\[X_n\xrightarrow{d}X\iff F_{X_n}(t)\longrightarrow F_X(t)\quad(t\in C(F_X))\]English content ke steps
Step 1: CDF Fₓₙ(t)=P(Xₙ≤t) hota hai. Har fixed continuity point t par n badhakar in probabilities ka limit check karte hain.
Step 2: Jis t par limiting CDF jump karta hai, definition wahan convergence demand nahi karti. C(Fₓ) ka matlab exactly continuity points ka set hai.
Step 3: Xₙ→ᵈX notation distributions ka limiting behaviour batati hai; ek hi experiment mein Xₙ aur X ki pointwise values equal ya close hona necessary nahi.
\(X \equiv 0\) w.p. 1.
\[ F_{X_n}(t) \longrightarrow F_X(t) \quad \forall t \in \mathbb{R} \setminus \{0\} \text{ (the set of all continuity pts of } X). \]Then \(X_n \xrightarrow{d} X\) as \(n \rightarrow \infty\).
Hinglish anuvaad
Xₙ ki value hamesha 1/n hai, aur X ki value hamesha 0 hai. Dikhana hai ki Xₙ distribution mein X ki taraf jaata hai.
📌 Formula: CDF continuity-point test
\[F_{X_n}(t)\longrightarrow F_X(t)\quad(t\ne0)\]English content ke steps
Step 1: t<0 par dono CDF zero hain. t>0 fix karo: n sufficiently bada ho to 1/n≤t, isliye Fₓₙ(t)=1 aur Fₓ(t)=1.
Step 2: t=0 par Fₓₙ(0)=0 har n ke liye, lekin Fₓ(0)=1. Yahan limit match nahi hota, phir bhi theorem apply hoti hai kyunki Fₓ zero par discontinuous hai.
Step 3: Isi example se samajho ki definition mein C(Fₓ) ki condition kyun likhi hai.
Let \(X_1, X_2, \dots\) be i.i.d random variables with finite mean \(\mu\) and variance \(\sigma^2\). Suppose \(X \sim \mathcal{N}(0,1)\) and
\[ S_n = \frac{X_1 + \dots + X_n - n\mu}{\sigma \sqrt{n}}. \]Then \(S_n \xrightarrow{d} X\).
\[ \implies F_{S_n}(t) \longrightarrow \Phi(t) \quad \text{for all } t \in \mathbb{R} \]where \(\Phi(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{t} e^{-\frac{x^2}{2}} dx\).
Remark: Suppose \(X_i \sim \mathcal{N}(\mu, \sigma^2)\). Then
\[ X_1 + X_2 + \dots + X_n \sim \mathcal{N}(n\mu, n\sigma^2). \] \[ \implies \frac{X_1 + X_2 + \dots + X_n - n\mu}{\sigma \sqrt{n}} \sim \mathcal{N}(0, 1). \]Hence the theorem holds in this case.
Hinglish anuvaad
i.i.d. X₁,X₂,… ka mean μ aur variance σ² finite hai. Sum mein se nμ ghata kar σ√n se divide karne par Sₙ banta hai; n badhne par Sₙ standard normal ke distribution ke paas aata hai.
📌 Formula: Central Limit Theorem
\[S_n=\frac{\sum_{i=1}^nX_i-n\mu}{\sigma\sqrt n}\xrightarrow{d}\mathcal N(0,1)\quad(\sigma>0)\]English content ke steps
Step 1: Sum ka mean nμ hai, isliye nμ subtraction center karta hai. Independent terms ke sum ki variance nσ² hai, isliye standard deviation σ√n se divide karke variance scale 1 hota hai.
Step 2: Φ(t) standard normal CDF hai. Limit mein P(Sₙ≤t)→Φ(t) for every t, kyunki standard normal CDF har jagah continuous hai.
Step 3: Remark mein agar har Xᵢ khud normal hai, to unka sum exactly N(nμ,nσ²) hai. Is special case mein Sₙ har n par N(0,1) hai; CLT ka limit immediately mil jata hai.
Step 4: σ=0 mein denominator zero ho jayega, isliye nondegenerate CLT mein σ>0 maana jaata hai.
We assume \(X_1, X_2, \dots\) are iid random variables s.t. moment generating fn. exists.
Let \(X \sim \mathcal{N}(0,1)\). Then
We show that \(M_{S_n}(t) \rightarrow M_X(t) = e^{t^2/2}\) as \(n \rightarrow \infty\).
\(\implies \log M_{S_n}(t) \rightarrow \frac{t^2}{2}\) as \(n \rightarrow \infty\).
Without loss of generality, assume that \(\mu = 0\) and \(\sigma^2 = 1\). Then \(S_n = \frac{X_1 + \dots + X_n}{\sqrt{n}}\).
\begin{align*} M_{S_n}(t) &= E\left[e^{\frac{t}{\sqrt{n}}(X_1 + \dots + X_n)}\right] = \prod_{i=1}^{n} E\left[e^{\frac{t X_i}{\sqrt{n}}}\right] \\ &= \left(E\left[e^{\frac{t X_1}{\sqrt{n}}}\right]\right)^n = \left(M_{X_1}\left(\frac{t}{\sqrt{n}}\right)\right)^n. \end{align*}Let \(M(t) = M_{X_1}(t)\) and \(L(t) = \log M(t)\). Note \(M(0) = 1\).
Now we have
Using L'Hôpital's rule:
\begin{align*} \lim_{n \rightarrow \infty} \log M_{S_n}(t) &= \lim_{n \rightarrow \infty} n \log M\left(\frac{t}{\sqrt{n}}\right) = \lim_{n \rightarrow \infty} \frac{L(t/\sqrt{n})}{1/n} \\ &= \lim_{n \rightarrow \infty} \frac{L'(t/\sqrt{n}) \cdot t \cdot (-\frac{1}{2} n^{-3/2})}{-n^{-2}} \quad \text{(by L'H\^opital rule)} \\ &= \lim_{n \rightarrow \infty} \frac{L'(t/\sqrt{n}) \cdot t}{2 n^{-1/2}} \\ &= \lim_{n \rightarrow \infty} \frac{L''(t/\sqrt{n}) \cdot t^2 \cdot (-\frac{1}{2} n^{-3/2})}{-n^{-3/2}} \quad \text{(by L'H\^opital rule)} \\ &= \lim_{n \rightarrow \infty} L''(t/\sqrt{n}) \frac{t^2}{2} \\ &= L''(0) \frac{t^2}{2} = \frac{t^2}{2}. \end{align*}\(\implies M_{S_n}(t) \rightarrow e^{t^2/2}\) as \(n \rightarrow \infty\).
Hence, \(S_n \xrightarrow{d} \mathcal{N}(0,1)\).
Hinglish anuvaad
Yeh CLT ka MGF se proof sketch hai, jab original variables ki MGF zero ke aas-paas exist kare. Pehle standard normal ki MGF nikaalte hain, phir standardized sum ki MGF ka limit usse compare karte hain.
📌 Formula: MGF proof ka limiting step
\[M_{S_n}(t)=\left[M_{X_1}\!\left(\frac{t}{\sqrt n}\right)\right]^n\longrightarrow e^{t^2/2}\]English content ke steps
Step 1: Normal MGF integral mein exponent tx−x²/2 ko −(x−t)²/2+t²/2 mein complete square kiya. Shifted normal density ka integral 1 hai, toh M_N(t)=e^(t²/2).
Step 2: μ=0 aur σ²=1 maan kar proof simpler hota hai; general case mein pehle variables ko (Xᵢ−μ)/σ se standardize karo.
Step 3: Independent Xᵢ ke exponential product ki expectation factor hoti hai. Identical distribution se sab factors M_X₁(t/√n) same hain, isliye n-th power aati hai.
Step 4: L(u)=log M_X₁(u) par L(0)=0, L′(0)=E[X₁]=0 aur L″(0)=Var(X₁)=1. Zero ke paas Taylor expansion L(u)=u²/2+o(u²) deta hai.
Step 5: u=t/√n rakhne par nL(t/√n)→t²/2. English proof mein isi limit ko L’Hôpital se do baar nikaala gaya hai. Exponential lene par MGF limit e^(t²/2) aur standard MGF uniqueness se convergence in distribution.
Step 6: MGF existence ki extra condition sirf is proof technique ke liye hai; CLT ka theorem finite variance ke saath bhi true hai jab original MGF na exist kare.
Toss a fair coin 100 times. Suppose \(X\) denotes the # Heads appeared. Find:
(1) \(P(\{40 \le X \le 60\})\) (2) \(P(\{X \ge 80\})\).
Solution: (use CLT)
\begin{align*} (1) \quad P(\{40 \le X \le 60\}) &= P(\{-10 \le X - 50 \le 10\}) \quad (\text{Here } n\mu = 100 \times \frac{1}{2} = 50) \\ &= P\left(\left\{-\frac{10}{5} \le \frac{X - 50}{5} \le \frac{10}{5}\right\}\right) \quad (\sigma^2 = 100 \cdot p(1-p) = 25 \implies \sigma = 5) \\ &\cong P(\{-2 \le Y \le 2\}) \quad \text{where } Y \sim \mathcal{N}(0,1) \\ &= \Phi(2) - \Phi(-2) \end{align*}where \(\Phi(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{t} e^{-\frac{x^2}{2}} dx\).
\begin{align*} (2) \quad P(\{X \ge 80\}) &= P\left(\left\{\frac{X - 50}{5} \ge \frac{30}{5}\right\}\right) \\ &\overset{CLT}{\cong} P(\{Y \ge 6\}) \\ &= 1 - \Phi(6). \end{align*}Hinglish anuvaad
Fair coin 100 baar toss hota hai. X Heads ki count hai. 40 se 60 Heads aur kam se kam 80 Heads ke chances ko CLT se approximately nikaalna hai.
📌 Formula: Binomial ka normal approximation
\[X\sim\operatorname{Bin}(100,1/2),\quad Z=\frac{X-50}{5}\approx\mathcal N(0,1)\]English content ke steps
Step 1: X ka mean np=50 aur variance np(1−p)=25 hai; standard deviation 5. X se 50 ghata kar 5 se divide karna standardization hai.
Step 2: (1) 40≤X≤60 ko −2≤Z≤2 likho. Normal approximation se P≈Φ(2)−Φ(−2), jo roughly 0.9545 hai.
Step 3: (2) X≥80 ko Z≥6 likho. Source ka CLT estimate 1−Φ(6) hai; itni door tail par normal approximation ka relative error kaafi ho sakta hai. Exact chance binomial tail sum se milega.
Step 4: Ye approximations hain, exact equalities nahi. Discrete binomial ko continuous normal se approximate karte waqt continuity correction laga sakte hain: part (1) mein 39.5 se 60.5 aur part (2) mein 79.5 boundary.
There are random variables whose all moments exist but moment generating fn. does not exist.
Example 1:
\(X \sim \mathcal{N}(0,1)\). Consider \(Y = X^4\). Clearly all moments exist, because \(E[Y^k] = E[X^{4k}]\) exists for all \(k \ge 1\).
But \(E[e^{tY}] = E[e^{tX^4}]\) does not exist for any \(t \neq 0\).
Example 2: Let \(X\) be a Pareto random variable with pdf, for \(\alpha > 0\),
\[ f_X(x) = \begin{cases} \frac{\alpha}{x^{\alpha+1}} & \text{if } x \ge 1 \\ 0 & \text{otherwise} \end{cases} \]Note: \(\int_{1}^{\infty} \frac{\alpha}{x^{\alpha+1}} dx = \left[ -\frac{\alpha}{\alpha x^\alpha} \right]_1^\infty = 1\).
\[ E[X^t] = \int_{1}^{\infty} \frac{\alpha}{x^{(\alpha-t)+1}} dx = \begin{cases} \frac{\alpha}{\alpha-t} & \text{if } t < \alpha \\ \text{Not exists} & \text{if } t \ge \alpha \end{cases} \]Thus the \(t\)-moment of \(X\) exists if \(t < \alpha\), otherwise does not exist.
In particular, Variance of \(X\) does not exist if \(\alpha \le 2\).
Also, mean of \(X\) does not exist if \(\alpha \le 1\).
Hinglish anuvaad
Is remark mein dekhenge ki moments aur MGF same baat nahi hain. Y=X⁴ mein saare integer moments finite hain; Pareto example mein kaunse moments finite hain, woh α se decide hota hai.
📌 Formula: Moments versus MGF
\[E[Y^k]=E[X^{4k}],\qquad E[\operatorname{Pareto}(\alpha)^t]=\frac\alpha{\alpha-t}\ (t<\alpha)\]English content ke steps
Step 1: Example 1: Standard normal X ke har polynomial moment finite hain, isliye Y=X⁴ ke E[Yᵏ]=E[X⁴ᵏ] bhi finite hain.
Step 2: Lekin positive t par E[e^(tX⁴)] ka integrand large |X| mein e^(tx⁴−x²/2) ki tarah grow karta hai; integral diverge karta hai. English line mein correction: t<0 par e^(tY)≤1 hai, toh expectation exist karti hai. Is Y ki MGF zero ke aas-paas dono taraf finite nahi hai.
Step 3: Example 2: Pareto density α/x^(α+1), x≥1, total area 1 hai. E[Xᵗ] ka integral α∫₁∞x^(t−α−1)dx hai; t<α par α/(α−t), t≥α par divergent.
Step 4: Mean t=1 par chahiye, isliye α>1; second moment t=2 par chahiye, isliye α>2. α≤2 par finite variance nahi hoti (mean missing ho to variance bhi defined nahi).
- (1) Suppose \(X\) and \(Y\) are two discrete random variables defined on the same probability space. The conditional mass function (pmf) of \(X\) given \(Y=y\) is defined by
\[
p_{X \mid Y}(x \mid y) = P(\{X=x \mid Y=y\}) = \frac{P(\{X=x, Y=y\})}{P(\{Y=y\})} = \frac{p_{X,Y}(x,y)}{p_Y(y)}
\]
for all values of \(y\) such that \(p_Y(y) > 0\).
Recall: \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\). - (2) The conditional probability distribution of \(X\) given \(Y=y\) is defined as \[ F_{X \mid Y}(x \mid y) = P(\{X \le x \mid Y=y\}) = \sum_{a \le x} p_{X \mid Y}(a \mid y). \]
Hinglish anuvaad
X,Y discrete random variables hain. Y=y already pata ho aur p_Y(y)>0 ho, toh X=x ka conditional PMF joint probability ko Y=y ke chance se divide karke milta hai. Conditional CDF in masses ka cumulative sum hai.
📌 Formula: Conditional PMF / CDF
\[p_{X\mid Y}(x\mid y)=\frac{p_{X,Y}(x,y)}{p_Y(y)},\qquad F_{X\mid Y}(a\mid y)=\sum_{x\le a}p_{X\mid Y}(x\mid y)\]English content ke steps
Step 1: (1) P(A|B)=P(A∩B)/P(B) mein A={X=x}, B={Y=y} rakho. Numerator joint PMF, denominator marginal PMF ban jaata hai.
Step 2: Denominator p_Y(y)>0 required hai. Fixed y par sab x ke conditional PMFs ka sum 1 hota hai.
Step 3: (2) X≤a ke liye X ki har possible value x≤a ke mutually exclusive conditional masses jodte hain. Isi se conditional CDF banta hai.
Suppose \(X \sim Poi(\lambda_1)\) and \(Y \sim Poi(\lambda_2)\) are independent R.V.s. Find the conditional mass fn. of \(X\) given \(X+Y=n\).
Solution:
\(X \sim Poi(\lambda_1)\) and \(Y \sim Poi(\lambda_2) \implies X+Y \sim Poi(\lambda_1 + \lambda_2)\).
\(p_{X \mid X+Y}(k \mid n) = P(\{X=k \mid X+Y=n\}) = 0\) if \(k > n\).
Suppose \(k \le n\).
Thus, \(X\) given \(X+Y=n\) is a \(Bin(n, \frac{\lambda_1}{\lambda_1+\lambda_2})\) random variable.
Hinglish anuvaad
X~Poi(λ₁), Y~Poi(λ₂) independent hain. Total X+Y=n pata ho, toh X ki conditional distribution nikaalni hai.
📌 Formula: Poisson split becomes Binomial
\[X\mid(X+Y=n)\sim\operatorname{Bin}\!\left(n,\frac{\lambda_1}{\lambda_1+\lambda_2}\right)\]English content ke steps
Step 1: Total n diya hai; X=k hoga toh Y=n−k hi ho sakta hai. k ko 0 se n tak hona chahiye; k>n ya k<0 par probability zero hai.
Step 2: Conditional probability mein numerator P(X=k,Y=n−k) aur denominator P(X+Y=n) rakho. Independence se numerator individual Poisson masses ka product hai.
Step 3: Denominator Poi(λ₁+λ₂) ka n-th mass hai. Exponential factors cancel, factorial ratio C(n,k) banta hai, aur λ₁/(λ₁+λ₂) success probability banti hai.
Step 4: Final PMF exactly Bin(n, λ₁/(λ₁+λ₂)) ka hai; λ₁+λ₂>0 aur P(X+Y=n)>0 maana gaya hai.
- (1) \(P(X=x) = \sum_y P(X=x, Y=y) = \sum_y p_{X \mid Y}(x \mid y) p_Y(y)\) (by def).
- (2) Suppose \(X\) and \(Y\) are independent, then \[ p_{X \mid Y}(x \mid y) = P(X=x \mid Y=y) = \frac{P(X=x, Y=y)}{P(Y=y)} = \frac{P(X=x)P(Y=y)}{P(Y=y)} = P(X=x). \]
Hinglish anuvaad
Pehla remark unconditional X=x ko Y ke sab cases mein divide karke sum karta hai. Doosra remark kehta hai ki X,Y independent hon toh Y=y pata hone se X ki PMF nahi badalti.
📌 Formula: Total probability / independence
\[p_X(x)=\sum_y p_{X\mid Y}(x\mid y)p_Y(y),\qquad X\perp Y\Rightarrow p_{X\mid Y}(x\mid y)=p_X(x)\]English content ke steps
Step 1: (1) Har y ke liye p_{X|Y}(x|y)p_Y(y)=p_{X,Y}(x,y). Saare y par sum lene se joint table ki row sum p_X(x) milti hai.
Step 2: (2) Independence mein p_{X,Y}(x,y)=p_X(x)p_Y(y). Conditional PMF formula mein substitute karke positive p_Y(y) cancel karo.
Step 3: Isliye conditional aur unconditional distribution match karti hain jab X aur Y independent hon.
Let \(X \sim Bin(n,p)\) and \(Y \sim Bin(m,p)\) be two independent random variables. Find the distribution of \(X+Y\).
Solution:
Suppose \(Z = X+Y\). Then
[Note: \((x+1)^n (x+1)^m = (x+1)^{n+m}\).
\((\sum \binom{n}{q} x^q) (\sum \binom{m}{j} x^j) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k\). Compare the coefficient of \(x^k\): \(\sum_{j=0}^{m} \binom{n}{k-j}\binom{m}{j} = \binom{n+m}{k}\). Assume \(\binom{n}{r} = 0\) if \(r<0\) or \(r>n\)].
Thus, \(X+Y \sim Bin(n+m, p)\).
Hinglish anuvaad
X~Bin(n,p), Y~Bin(m,p) independent hain. Inka sum Z=X+Y ka distribution find karna hai.
📌 Formula: Binomial convolution / Vandermonde identity
\[X+Y\sim\operatorname{Bin}(n+m,p),\quad\sum_j\binom n{k-j}\binom mj=\binom{n+m}k\]English content ke steps
Step 1: Z=k ho sakta hai jab Y=j aur X=k−j, jitne valid j hain. Total probability se in disjoint cases ka sum likho.
Step 2: Independent hone se P(X=k−j,Y=j)=P(X=k−j)P(Y=j). Binomial PMFs substitute karke p ke exponents k aur (1−p) ke n+m−k collect karo.
Step 3: Remaining combinations ki sum Vandermonde identity se C(n+m,k) hai. English bracket mein (1+x)^n(1+x)^m=(1+x)^(n+m) ke xᵏ coefficients compare karke identity derive ki.
Step 4: Intuition: n trials aur m independent trials, same success chance p ke saath, total n+m trials banate hain.
Let \(X\) and \(Y\) be two discrete R.V.s.
\[ E[X \mid Y=y] = \sum_x x \, p_{X \mid Y}(x \mid y) = \sum_x x \frac{p_{X,Y}(x,y)}{p_Y(y)}. \]Hinglish anuvaad
Discrete X ki conditional expectation given Y=y, X ki possible values ko conditional PMF se weight karke sum karna hai.
📌 Formula: Discrete conditional expectation
\[E[X\mid Y=y]=\sum_x x\,p_{X\mid Y}(x\mid y)=\sum_xx\,\frac{p_{X,Y}(x,y)}{p_Y(y)}\]English content ke steps
Step 1: Ordinary E[X]=Σx·p_X(x) tha. Y=y pata hone par p_X ko conditional p_{X|Y} se replace kiya gaya hai.
Step 2: Formula valid hai jab p_Y(y)>0 aur relevant expectation finite/defined ho. Har y ke liye alag conditional mean mil sakta hai.
Suppose \(X \sim Poi(\lambda_1)\) and \(Y \sim Poi(\lambda_2)\) are two independent random variables. Find the conditional expected value of \(X\) given \(X+Y=n\).
Solution:
\begin{align*} E[X \mid X+Y=n] &= \sum_{k=0}^{n} k \cdot P(X=k \mid X+Y=n) \\ &\overset{ex. 1}{=} \sum_{k=0}^{n} k \cdot \binom{n}{k} \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^k \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n-k} \\ &= n \cdot \frac{\lambda_1}{\lambda_1+\lambda_2} \sum_{k=1}^{n} \binom{n-1}{k-1} \left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^{k-1} \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{(n-1)-(k-1)} \\ &= n \cdot \frac{\lambda_1}{\lambda_1+\lambda_2} \cdot 1 = n \frac{\lambda_1}{\lambda_1+\lambda_2}. \end{align*}Hinglish anuvaad
Do independent Poisson counts X aur Y ka total n given hai. Is information ke baad X ka expected count poochha hai.
📌 Formula: Conditional Poisson split ka mean
\[E[X\mid X+Y=n]=n\frac{\lambda_1}{\lambda_1+\lambda_2}\]English content ke steps
Step 1: Pichhle example mein X given total n ka distribution Bin(n,q) mila, jahan q=λ₁/(λ₁+λ₂). Isliye expected value nq honi chahiye.
Step 2: English sum mein k·C(n,k)=n·C(n−1,k−1) use karke nq bahar nikala. Bachi n−1 trials ki binomial probabilities ka sum 1 hai.
Step 3: Intuition: Total n arrivals mein har arrival ka X source se aane ka expected fraction q hai.
\(\sum_y E[X \mid Y=y] P(Y=y) = E[X]\).
Proof:
\begin{align*} \sum_y E[X \mid Y=y] P(Y=y) &= \sum_y \sum_x x \, p_{X \mid Y}(x \mid y) P(Y=y) \\ &= \sum_y \sum_x x \, p_{X,Y}(x,y) \\ &= \sum_x x \sum_y p_{X,Y}(x,y) \\ &= \sum_x x \, p_X(x) = E[X]. \end{align*}Hinglish anuvaad
Conditional expectations ko conditioning variable Y ke probabilities se average karne par original E[X] wapas milta hai. Isko total expectation ya tower property kehte hain.
📌 Formula: Law of total expectation
\[E[X]=\sum_yE[X\mid Y=y]P(Y=y)\]English content ke steps
Step 1: Proof ki first line conditional mean Σx·p_{X|Y}(x|y) ko substitute karti hai.
Step 2: p_{X|Y}(x|y)P(Y=y)=p_{X,Y}(x,y) hone par expression double sum mein badalta hai.
Step 3: Sum ka order badal kar fixed x ke liye y par sum lo: Σy p_{X,Y}(x,y)=p_X(x). Ab Σx x p_X(x)=E[X] milta hai.
A miner is trapped in a mine containing 3 doors. The first door leads to a tunnel that will take him to safety after 3 hours of travel. The 2nd and 3rd doors lead to a tunnel that will return him to the mine after 5 & 7 hours (respectively). If we assume that the miner is at all times equally likely to choose any one of the doors, what is the expected length of time until he reaches safety?
Solution:
Suppose \(X =\) the amount of time until the miner reaches safety.
\(Y =\) the door he initially chooses.
\(\implies E[X] - \frac{2}{3}E[X] = 5 \implies \frac{1}{3}E[X] = 5 \implies E[X] = 15\).
Hinglish anuvaad
Miner teen doors mein se har baar equally choose karta hai. Door 1 se 3 hours mein exit; doors 2 aur 3 se 5 aur 7 hours ke baad mine mein wapas. Safety tak expected total time nikaalna hai.
📌 Formula: First-step analysis / total expectation
\[E[X]=\frac13\bigl(3+(5+E[X])+(7+E[X])\bigr)=15\ \text{hours}\]English content ke steps
Step 1: Y first chosen door aur X total time hai. Y=1 par X=3. Y=2 par 5 hours waste karke same starting state mein lautte hain: conditional mean 5+E[X].
Step 2: Y=3 par 7 hours ke baad wahi state hai: conditional mean 7+E[X]. Har door 1/3 chance se choose hota hai, so in teen conditional means ka average lo.
Step 3: Equation E[X]=(15+2E[X])/3 se E[X]/3=5 aur E[X]=15 hours. Wapas aane ke baad choices same rule se hoti hain, ye assumption recurrence ko justify karti hai.
| A | |||
|---|---|---|---|
| Confess | Silent | ||
| B | Confess | (5, 5) | (0, 10) |
| Silent | (10, 0) | (1, 1) | |
(Penalties: A, B)
\(X = \text{time in jail given confess}\).
Conclusion: Prisoner will confess irrespective of other's decision (as \(2.5 < 5.5\)).
Hinglish anuvaad
Table mein A aur B dono Confess ya Silent choose karte hain. Ordered pair (A ki penalty, B ki penalty) hai. Given equal 1/2 choice assumption mein A ke expected jail times compare kiye gaye hain.
📌 Formula: Conditional expectation / payoff comparison
\[E[\text{A penalty}\mid\text{Confess}]=2.5,\quad E[\text{A penalty}\mid\text{Silent}]=5.5\]English content ke steps
Step 1: Table padho: B Confess ho toh A Confess par 5, A Silent par 10; B Silent ho toh A Confess par 0, A Silent par 1. Dono columns mein confess A ki penalty kam karta hai.
Step 2: Agar B Confess aur Silent ko 1/2–1/2 se choose kare, confession mean (5+0)/2=2.5 aur silence mean (10+1)/2=5.5 hai.
Step 3: English solution mein P(Y₂=1)=1/2 assume kiya gaya hai; question ke table se yeh probability khud nahi milti. Fir bhi pairwise comparison se Confess dominant action hai, B ki choice probabilities jaane bina.
Step 4: Isi symmetry se B ke liye bhi confession individually better hai, although both Silent (1,1) both Confess (5,5) se better outcome hota hai.
- (1) (Conditional pdf) Suppose \(X\) and \(Y\) be two continuous random variables with joint density \(f_{X,Y}(x,y)\). Then the conditional probability density function of \(X\) given that \(Y=y\) is defined, for all values of \(y\) such that \(f_Y(y) > 0\), by \[ f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}. \]
- (2) Conditional distribution function of \(X\) given \(Y=y\) is defined as \[ F_{X \mid Y}(a \mid y) := P(\{X \le a \mid Y=y\}) = \int_{-\infty}^{a} f_{X \mid Y}(x \mid y) dx. \]
- (3) Conditional expectation of \(X\) given \(Y=y\) is defined by \[ E[X \mid Y=y] = \int_{-\infty}^{\infty} x f_{X \mid Y}(x \mid y) dx. \]
Hinglish anuvaad
Joint density wale continuous X,Y mein Y=y given hone par conditional density, conditional CDF aur conditional mean ki definitions di hain.
📌 Formula: Conditional PDF / CDF / expectation
\[f_{X\mid Y}(x\mid y)=\frac{f_{X,Y}(x,y)}{f_Y(y)},\quad E[X\mid Y=y]=\int x f_{X\mid Y}(x\mid y)\,dx\]English content ke steps
Step 1: (1) Joint density ko fixed y par marginal f_Y(y) se normalize karte hain. f_Y(y)>0 chahiye; x par conditional density ka integral 1 hota hai.
Step 2: (2) −∞ se a tak conditional density integrate karne par conditional CDF F_{X|Y}(a|y) milta hai.
Step 3: (3) x ko conditional density se weight karke sab x par integrate karo, conditional expected value milti hai.
Step 4: Continuous Y mein usually P(Y=y)=0 hota hai; isliye discrete conditional-PMF wala 0/0 ratio direct nahi laga sakte. Density ratio appropriate formula hai.
Let \(X\) be a continuous random variable with pdf \(f(x)\).
Note: \(f(x) \neq P(\{X=x\})\) (\(=0\), in fact).
Note:
Hinglish anuvaad
Continuous random variable ki pdf f(a), exact point probability P(X=a) nahi hoti. f(a) ko a ke bahut paas ke chhote interval ki probability per unit length ka limit samjho.
📌 Formula: Density as local probability per unit length
\[f(a)=\lim_{\varepsilon\to0}\frac{P(a-\varepsilonEnglish content ke steps
Step 1: Interval ki width 2ε hai. Us interval ka probability F(a+ε)−F(a−ε) hai; ise width se divide karke local average density milti hai.
Step 2: ε→0 par F a par differentiable ho to symmetric difference quotient F′(a)=f(a) tak jaata hai.
Step 3: Single point interval ki width zero hoti hai aur continuous distribution mein P(X=a)=0. f(a) 1 se bada bhi ho sakta hai: density aur probability alag quantities hain.
The joint density function of \(X\) and \(Y\) is given by
\[ f(x,y) = \begin{cases} x e^{-x(y+1)} & x > 0, y > 0 \\ 0 & \text{otherwise} \end{cases} \]Find (1) the conditional density of \(X\) given \(Y=y\).
(2) Find distribution of \(X\) given \(Y=y\).
(3) Find expectation of \(X\) given \(Y=y\).
Solution:
\begin{align*} f_Y(y) &= \int_{0}^{\infty} f(x,y)dx = \int_{0}^{\infty} x e^{-x(y+1)}dx \\ &= \left[ -x \frac{e^{-x(y+1)}}{(y+1)} \right]_0^\infty + \int_{0}^{\infty} \frac{e^{-x(y+1)}}{(y+1)} dx \\ &= \frac{1}{y+1} \int_{0}^{\infty} e^{-x(y+1)} dx \\ &= \frac{1}{y+1} \left[ -\frac{e^{-x(y+1)}}{y+1} \right]_0^\infty \\ &= \frac{1}{(y+1)^2} \quad \text{for } y > 0. \end{align*}Note: \(\int_{0}^{\infty} f_Y(y)dy = \int_{0}^{\infty} \frac{1}{(y+1)^2} dy = 1 \implies \int_{0}^{\infty} \int_{0}^{\infty} f(x,y)dxdy = 1\).
Therefore,
- (1) \[ f_{X \mid Y}(x \mid y) = \frac{f(x,y)}{f_Y(y)} = \frac{x e^{-x(y+1)}}{1/(y+1)^2} = x(y+1)^2 e^{-x(y+1)} \quad \text{for } x > 0, y > 0. \]
- (2) \[ F_{X \mid Y}(a \mid y) = \int_{0}^{a} \frac{f(x,y)}{f_Y(y)} dx = \frac{\int_{0}^{a} f(x,y) dx}{f_Y(y)}. \]
- (3) \begin{align*} E[X \mid Y=y] &= \int_{0}^{\infty} x f_{X \mid Y}(x \mid y) dx = \int_{0}^{\infty} x \left( x(y+1)^2 e^{-x(y+1)} \right) dx \\ &= (1+y)^2 \int_{0}^{\infty} x^2 e^{-x(y+1)} dx. \end{align*} Let \(x(y+1) = t \implies dx = \frac{dt}{1+y}\). \begin{align*} E[X \mid Y=y] &= (1+y)^2 \int_{0}^{\infty} \frac{t^2}{(1+y)^2} e^{-t} \frac{dt}{1+y} \\ &= \frac{1}{1+y} \int_{0}^{\infty} t^2 e^{-t} dt \\ &= \frac{\Gamma(3)}{1+y} = \frac{2}{1+y}. \end{align*}
Hinglish anuvaad
Joint density f(x,y)=x e^(−x(y+1)) for x>0,y>0 hai. Y=y par X ki conditional density, uska conditional distribution aur mean nikaalna hai.
📌 Formula: Marginalise, divide, integrate
\[f_{X\mid Y}(x\mid y)=x(1+y)^2e^{-(1+y)x},\qquad E[X\mid Y=y]=\frac2{1+y}\]English content ke steps
Step 1: Sabse pehle f_Y(y)=∫₀∞x e^(−(1+y)x)dx=1/(1+y)². English solution mein integration by parts ka boundary term zero hota hai; y>0 par total marginal area 1 check hota hai.
Step 2: (1) Conditional PDF joint/marginal ratio hai: x e^(−(1+y)x) ko 1/(1+y)² se divide karke x(1+y)²e^(−(1+y)x), x>0.
Step 3: (2) Conditional CDF ke liye 0 se a tak density integrate karo. Source integral par rukta hai; evaluate karoge toh a≤0 par 0, aur a>0 par 1−e^(−(1+y)a)[1+(1+y)a] milta hai.
Step 4: (3) Mean ke integral mein x se ek additional factor aata hai, isliye x² integrand banta hai. t=(1+y)x substitution se Γ(3)=2! = 2; result 2/(1+y).
Let \(Z_0 = 1\) and \(Z_n\) denote the size of the population at time \(n\), and let \(X_{n,i}\) denote the number of children of \(i\)-th member in \((n-1)\)-th step.
Where \(\{X_{n,i} : i=1, \dots, Z_{n-1}, n \ge 1\}\) are iid R.V.s. Then
Find \(E[Z_{n+1}]\).
Solution:
\(Z_0 = 1\)
\(Z_1 = X_{1,1}\)
\(Z_2 = X_{2,1} + \dots + X_{2,Z_1}\)
Remark: Suppose \(\mu < 1\). Then
- (1) \(P(Z_n \ge 1) \le E[Z_n] = \mu^n \rightarrow 0\) as \(n \rightarrow \infty\) (Markov's inequality).
- (2) In fact, if \(\mu < 1\), \(P(\{Z_n > \epsilon\}) \rightarrow 0\) as \(n \rightarrow \infty\). \(\implies Z_n \xrightarrow{i.p.} 0\).
By Borel-Cantelli lemma, \(\mu < 1 \implies Z_n \xrightarrow{a.s.} 0\) as \(n \rightarrow \infty\).
Hinglish anuvaad
Z₀=1 initial population hai. Har generation n mein pichhli generation ke har member i ke children Xₙ,ᵢ hain. Zₙ un children ka sum hai; E[Zₙ₊₁] nikaalna hai.
📌 Formula: Branching recursion + total expectation
\[E[Z_n]=\mu E[Z_{n-1}]=\mu^n,\qquad E[Z_{n+1}]=\mu^{n+1}\]English content ke steps
Step 1: Agar Zₙ₋₁=k known hai, toh exactly k parents hain. Har parent ka expected offspring μ hone se linearity ke through conditional expected population kμ hai.
Step 2: Ab k ke sab possible values par average lo: E[Zₙ]=Σk kμ P(Zₙ₋₁=k)=μE[Zₙ₋₁]. Z₀=1 se repeatedly E[Zₙ]=μⁿ aur asked answer E[Zₙ₊₁]=μⁿ⁺¹.
Step 3: μ<1 par Markov: P(Zₙ≥1)≤E[Zₙ]=μⁿ→0. Kyunki no-immigration branching model mein zero population ke baad population zero rehti hai, extinction eventually almost surely hoti hai; source mein Borel–Cantelli route ka remark diya hai.
Step 4: For ε>0, Markov se P(Zₙ>ε)≤μⁿ/ε→0 bhi milta hai: Zₙ probability mein zero ki taraf jaata hai.
- (1) Let \(X \sim Geo(p)\). Then \(P(X > s+t \mid X > s) = P(X > t)\) (memoryless property).
Proof: \begin{align*} P(X > s+t \mid X > s) &= \frac{P(\{X > s+t\} \cap \{X > s\})}{P(X > s)} \\ &= \frac{P(X > s+t)}{P(X > s)} \\ &= \frac{(1-p)^{s+t}}{(1-p)^s} \\ &= (1-p)^t = P(X > t). \end{align*} Note: \(P(X > s) = \sum_{k=s+1}^{\infty} p(1-p)^{k-1} = p \frac{(1-p)^s}{1 - (1-p)} = (1-p)^s\). - (2) If \(X \sim \exp(\lambda)\) then \(P(X > s+t \mid X > s) = P(X > t)\).
Proof:
Note: \(P(X > s) = \int_{s}^{\infty} \lambda e^{-\lambda x}dx = e^{-\lambda s}\).
Therefore, \begin{align*} P(X > t+s \mid X > s) &= \frac{P(X > t+s, X > s)}{P(X > s)} \\ &= \frac{P(X > t+s)}{P(X > s)} \\ &= \frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} = P(X > t). \end{align*}
In fact, these are the only random variables which have the memoryless property.
Hinglish anuvaad
Geometric waiting time aur exponential waiting time dono memoryless hain: s time/trials wait karne ke baad aur t wait karne ka chance initial t wait ke chance jaisa hi rehta hai.
📌 Formula: Memoryless survival probabilities
\[P(X>s+t\mid X>s)=\frac{P(X>s+t)}{P(X>s)}=P(X>t)\]English content ke steps
Step 1: (1) Geometric X first success ki trial index hai. s trials tak failure ka chance (1−p)^s hai. Ratio (1−p)^(s+t)/(1−p)^s=(1−p)^t; proof mein geometric-series sum se survival nikala hai.
Step 2: (2) Exponential X ki survival P(X>s)=∫ₛ∞λe^(−λx)dx=e^(−λs) hai. Ratio e^(−λ(s+t))/e^(−λs)=e^(−λt).
Step 3: Geometric mein s,t nonnegative integers aur 0
Step 4: Source ka last sentence usual nondegenerate, suitable discrete/continuous waiting-time distributions ke context mein samjho: geometric aur exponential canonical memoryless families hain.