Homework 3: Probability and its Application (MTH432/532)

English solution

Set-up

Write the two rolls as an ordered pair \((D_1,D_2)\), where each of \(D_1,D_2\) belongs to \(\{1,2,3,4,5,6\}\). We only need the range of each random variable, not the probability of each value.

(a) Maximum of the two rolls

Let \(M=\max(D_1,D_2)\). The smallest possible maximum is \(1\), obtained from \((1,1)\), and the largest is \(6\). Every integer in between occurs, for example from \((k,k)\). Hence

\[M\in\{1,2,3,4,5,6\}.\]

(b) Minimum of the two rolls

Let \(m=\min(D_1,D_2)\). Its smallest possible value is \(1\), its largest possible value is \(6\), and every intermediate value can occur. Therefore

\[m\in\{1,2,3,4,5,6\}.\]

(c) Sum of the two rolls

Let \(S=D_1+D_2\). The smallest sum is \(1+1=2\), and the largest is \(6+6=12\). Every integer between them is possible:

\[S\in\{2,3,4,5,6,7,8,9,10,11,12\}.\]

(d) First roll minus second roll

Let \(R=D_1-D_2\). The minimum is \(1-6=-5\), while the maximum is \(6-1=5\). Every integer between these extremes is attainable, so

\[R\in\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}.\]

Final answers

• (a) \(\{1,2,3,4,5,6\}\)
• (b) \(\{1,2,3,4,5,6\}\)
• (c) \(\{2,3,\ldots,12\}\)
• (d) \(\{-5,-4,\ldots,5\}\)

English solution

Main idea: probability equals jump size

For a discrete random variable, the CDF is \(F_X(x)=P(X\le x)\). The mass at a point \(a\) equals the jump of the CDF at \(a\):

\[p_X(a)=P(X=a)=F_X(a)-F_X(a^-),\]

where \(F_X(a^-)=\lim_{x\uparrow a}F_X(x)\). The given CDF jumps at \(0,1,2,3\), and \(3.5\).

Compute each jump

1. At \(x=0\):

   \[P(X=0)=\frac12-0=\frac12.\]

2. At \(x=1\):

   \[P(X=1)=\frac23-\frac12=\frac16.\]

3. At \(x=2\):

   \[P(X=2)=\frac45-\frac23=\frac{12-10}{15}=\frac2{15}.\]

4. At \(x=3\):

   \[P(X=3)=\frac9{10}-\frac45=\frac1{10}.\]

5. At \(x=3.5\):

   \[P(X=3.5)=1-\frac9{10}=\frac1{10}.\]

Probability mass function

\[ p_X(x)=P(X=x)= \begin{cases} \frac12, & x=0,\\ \frac16, & x=1,\\ \frac2{15}, & x=2,\\ \frac1{10}, & x=3,\\ \frac1{10}, & x=3.5,\\ 0, & \text{otherwise}. \end{cases} \]

Verification

Every mass is non-negative, and

\[\frac12+\frac16+\frac2{15}+\frac1{10}+\frac1{10}=\frac{15+5+4+3+3}{30}=1.\]

Therefore this is a valid PMF.

English solution

(a) Find the normalizing constant \(c\)

A probability density must integrate to 1:

\[ 1=\int_{-\infty}^{\infty}f_X(x)\,dx =c\int_0^2(4x-2x^2)\,dx. \]

Now

\[ \int_0^2(4x-2x^2)\,dx =\left[2x^2-\frac23x^3\right]_0^2 =8-\frac{16}{3}=\frac83. \]

Thus \(c\cdot\frac83=1\), so

\[\boxed{c=\frac38}.\]

(b) Find \(P(\frac12<X<\frac32)\)

For a continuous random variable, interval probability is the area under the density:

\[ P\left(\frac12<X<\frac32\right) =\frac38\int_{1/2}^{3/2}(4x-2x^2)\,dx. \]

An antiderivative after multiplying by \(3/8\) is

\[\frac34x^2-\frac14x^3.\]

Therefore

\[ P\left(\frac12<X<\frac32\right) =\left[\frac34x^2-\frac14x^3\right]_{1/2}^{3/2} =\frac{27}{32}-\frac5{32} =\boxed{\frac{11}{16}}. \]

(c) Find \(E[X]\)

\[ E[X]=\int_0^2x f_X(x)\,dx =\frac38\int_0^2(4x^2-2x^3)\,dx. \]

\[ E[X]=\frac38\left[\frac43x^3-\frac12x^4\right]_0^2 =\frac38\left(\frac{32}{3}-8\right) =\boxed{1}. \]

Find \(E[X^2]\) and the variance

\[ E[X^2]=\frac38\int_0^2(4x^3-2x^4)\,dx =\frac38\left[x^4-\frac25x^5\right]_0^2 =\frac38\left(16-\frac{64}{5}\right) =\frac65. \]

Using \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\),

\[ \operatorname{Var}(X)=\frac65-1^2=\boxed{\frac15}. \]

Symmetry check

The density satisfies \(f_X(1-u)=f_X(1+u)\), so it is symmetric about \(x=1\). Hence \(E[X]=1\) is consistent with the shape of the density.

Final answers

\[c=\frac38,\qquad P\left(\frac12<X<\frac32\right)=\frac{11}{16},\qquad E[X]=1,\qquad \operatorname{Var}(X)=\frac15.\]

English solution

(a) Geometric distribution

We use the common convention that \(X\) is the trial number of the first success:

\[P(X=k)=p(1-p)^{k-1},\qquad k=1,2,\ldots\]

Let \(q=1-p\). Its probability generating function is

\[ G_X(s)=E[s^X]=\sum_{k=1}^{\infty}s^k p q^{k-1} =\frac{ps}{1-qs}. \]

Since \(E[X]=G_X'(1)\) and \(E[X(X-1)]=G_X''(1)\),

\[E[X]=\frac1p,\qquad \operatorname{Var}(X)=\frac{1-p}{p^2}.\]

Convention note: If \(X\) counts failures before the first success, so its support is \(0,1,2,\ldots\), then \(E[X]=(1-p)/p\); the variance remains \((1-p)/p^2\).

(b) Poisson distribution

For \(X\sim\operatorname{Poisson}(\lambda)\),

\[P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots\]

The MGF is \(M_X(t)=\exp(\{\lambda(e^t-1)\})\). Differentiating at \(t=0\) gives

\[E[X]=\lambda,\qquad \operatorname{Var}(X)=\lambda.\]

(c) Normal distribution

The notation \(X\sim N(\mu,\sigma^2)\) already identifies its location and spread:

\[E[X]=\mu,\qquad \operatorname{Var}(X)=\sigma^2.\]

(d) Exponential distribution

Using the rate parameterization,

\[f_X(x)=\lambda e^{-\lambda x},\qquad x\ge0,\quad \lambda>0.\]

Direct integration, or differentiation of \(M_X(t)=\lambda/(\lambda-t)\), gives

\[E[X]=\frac1\lambda,\qquad \operatorname{Var}(X)=\frac1{\lambda^2}.\]

Final summary

Distribution	Mean	Variance
\(\operatorname{Geo}(p)\), support \(1,2,\ldots\)	\(1/p\)	\((1-p)/p^2\)
\(\operatorname{Poisson}(\lambda)\)	\(\lambda\)	\(\lambda\)
\(N(\mu,\sigma^2)\)	\(\mu\)	\(\sigma^2\)
\(\operatorname{Exp}(\lambda)\), rate \(\lambda\)	\(1/\lambda\)	\(1/\lambda^2\)

English solution

Definition of an MGF

For a random variable \(Z\), its moment generating function is

\[M_Z(t)=E[e^{tZ}],\]

for those \(t\) where the expectation is finite.

(a) MGF of a Binomial random variable

For \(X_n\sim\operatorname{Bin}(n,p)\),

\[ \begin{aligned} M_{X_n}(t) &=\sum_{k=0}^{n}e^{tk}\binom nk p^k(1-p)^{n-k}\\ &=\sum_{k=0}^{n}\binom nk(pe^t)^k(1-p)^{n-k}\\ &=(1-p+pe^t)^n, \end{aligned} \]

by the binomial theorem. Equivalently, \(M_{X_n}(t)=[1+p(e^t-1)]^n\).

MGF of a Poisson random variable

For \(Y\sim\operatorname{Poisson}(\lambda)\),

\[ \begin{aligned} M_Y(t) &=\sum_{k=0}^{\infty}e^{tk}e^{-\lambda}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!}\\ &=e^{-\lambda}e^{\lambda e^t} =\exp\{\lambda(e^t-1)\}. \end{aligned} \]

(b) Limit of the Binomial MGF

Here \(p\) must depend on \(n\); write it as \(p_n\). The assumption is \(np_n\to\lambda\), which also implies \(p_n\to0\). Put \(a=e^t-1\). Then

\[M_{X_n}(t)=(1+p_na)^n.\]

Take logarithms:

\[ \log M_{X_n}(t)=n\log(1+p_na). \]

Since \(p_n\to0\), use \(\log(1+u)=u+O(u^2)\) as \(u\to0\):

\[ n\log(1+p_na)=np_na+O(np_n^2a^2). \]

Now \(np_n\to\lambda\), and

\[np_n^2=(np_n)p_n\to\lambda\cdot0=0.\]

Therefore

\[ \log M_{X_n}(t)\to\lambda a=\lambda(e^t-1). \]

Exponentiating both sides gives

\[ \boxed{M_{X_n}(t)\to\exp\{\lambda(e^t-1)\}=M_Y(t)}. \]

Hence, by the stated MGF convergence theorem,

\[X_n\xrightarrow{d}Y\sim\operatorname{Poisson}(\lambda).\]

Special case \(p_n=\lambda/n\)

Then the limit is the familiar exponential limit:

\[ M_{X_n}(t)=\left(1+\frac{\lambda(e^t-1)}{n}\right)^n \longrightarrow e^{\lambda(e^t-1)}. \]

English solution

Understand the indicator

Fix one outcome and suppose the realized value of \(X\) is \(x\ge0\). Then

\[ I_{\{X>t\}}= \begin{cases} 1, & 0\le t<x,\\ 0, & t\ge x. \end{cases} \]

Thus the indicator simply cuts the infinite integral off at \(t=x\).

(a) Pointwise identity

For the realized value \(X=x\),

\[ \begin{aligned} \int_0^\infty I_{\{X>t\}}g'(t)\,dt &=\int_0^x g'(t)\,dt\\ &=g(x)-g(0)\\ &=g(x), \end{aligned} \]

where the second equality follows from the Fundamental Theorem of Calculus and the last uses \(g(0)=0\). Since this holds for every outcome,

\[ \boxed{g(X)=\int_0^\infty I_{\{X>t\}}g'(t)\,dt}. \]

(b) Take expectations and interchange expectation and integral

Starting from part (a),

\[ E[g(X)] =E\left[\int_0^\infty I_{\{X>t\}}g'(t)\,dt\right]. \]

Using Tonelli's theorem when \(g'(t)\ge0\), or Fubini's theorem whenever the integrand is absolutely integrable, we may interchange expectation and integration:

\[ E[g(X)]=\int_0^\infty E[I_{\{X>t\}}]g'(t)\,dt. \]

The expectation of an indicator equals the probability of its event:

\[ E[I_{\{X>t\}}]=1\cdot P(X>t)+0\cdot P(X\le t)=P(X>t). \]

Since \(P(X>t)=1-F_X(t)=\overline F_X(t)\),

\[ \boxed{E[g(X)]=\int_0^\infty \overline F_X(t)g'(t)\,dt}. \]

Important notation

The ordinary CDF is \(F_X(t)=P(X\le t)\). The bar denotes the survival or tail function:

\[\overline F_X(t)=1-F_X(t)=P(X>t).\]

Useful consequences

Choosing \(g(x)=x\), so \(g'(t)=1\), gives the tail-sum formula

\[E[X]=\int_0^\infty P(X>t)\,dt.\]

Choosing \(g(x)=x^r\) for \(r>0\), so \(g'(t)=rt^{r-1}\), gives

\[E[X^r]=r\int_0^\infty t^{r-1}P(X>t)\,dt,\]

whenever the relevant expectation/integral exists.

Technical condition

The interchange is automatically justified by Tonelli when \(g'\ge0\), because the integrand is then non-negative. For a general differentiable \(g\), one assumes the absolute-integrability condition

\[\int_0^\infty P(X>t)|g'(t)|\,dt<\infty,\]

which allows Fubini's theorem to be applied.

English recap

• Q1: max/min values are \(1\) to \(6\); sum is \(2\) to \(12\); difference is \(-5\) to \(5\).
• Q2: PMF values are the jump sizes of the CDF.
• Q3: normalize a density first; \(c=3/8\), middle-interval probability \(11/16\), mean \(1\), variance \(1/5\).
• Q4: memorize standard means and variances, but state the geometric/exponential convention.
• Q5: \(M_{\mathrm{Bin}}(t)=(1-p+pe^t)^n\), \(M_{\mathrm{Pois}}(t)=e^{\lambda(e^t-1)}\).
• Q6: an indicator converts \(\int_0^\infty\) into \(\int_0^X\), and its expectation is an event probability.