Random Variables and Distributions: Difficult Problems with Solutions

Solution:
The parameter of a Poisson distribution equals its mean. For a piece of [tex]4[/tex] square feet the mean number of defects is [tex]\lambda=2.5\times 4=10[/tex], so the number of defects [tex]X[/tex] satisfies
[tex]P(X=k)=e^{-10}\frac{10^k}{k!},\quad k=0, 1, 2, \ldots[/tex]
"Fewer than [tex]5[/tex]" means [tex]X\le 4[/tex]:
[tex]P(X\lt 5)=e^{-10}\left(1+10+\frac{10^2}{2!}+\frac{10^3}{3!}+\frac{10^4}{4!}\right)=e^{-10}\left(1+10+50+\frac{500}{3}+\frac{1250}{3}\right)[/tex]
[tex]P(X\lt 5)=\frac{1933}{3}e^{-10}\approx 0.029[/tex]

Solution:
Each bet wins with probability [tex]p=\frac{18}{38}=\frac{9}{19}[/tex] and loses with probability [tex]q=\frac{10}{19}[/tex], independently of the others. The player makes exactly [tex]9[/tex] bets if and only if the ninth bet is a win and among the first [tex]8[/tex] bets there are exactly [tex]3[/tex] wins and [tex]5[/tex] losses. Hence the number of bets [tex]X[/tex] has a negative binomial (Pascal) distribution and
[tex]P(X=9)=\binom{8}{3}p^3q^5\times p=\binom{8}{3}p^4q^5=56\left(\frac{9}{19}\right)^4\left(\frac{10}{19}\right)^5\approx 0.114[/tex]

Solution:
The marginal distributions are the row and column sums of the table:
[tex]P(X=0)=P(X=1)=0.5,\qquad P(Y=1)=P(Y=2)=0.3,\quad P(Y=3)=0.4[/tex]
Expected values and variances:
[tex]E(X)=0.5,\qquad D(X)=E(X^2)-(E(X))^2=0.5-0.25=0.25[/tex]
[tex]E(Y)=1\times 0.3+2\times 0.3+3\times 0.4=2.1,\qquad E(Y^2)=1\times 0.3+4\times 0.3+9\times 0.4=5.1[/tex]
[tex]D(Y)=E(Y^2)-(E(Y))^2=5.1-4.41=0.69[/tex]
Since [tex]XY=0[/tex] when [tex]X=0[/tex], only the row [tex]X=1[/tex] contributes to [tex]E(XY)[/tex]:
[tex]E(XY)=1\times 0.2+2\times 0.2+3\times 0.1=0.9,\qquad \operatorname{cov}(X,Y)=E(XY)-E(X)E(Y)=0.9-1.05=-0.15[/tex]
Hence the correlation coefficient is
[tex]\rho=\frac{\operatorname{cov}(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}=\frac{-0.15}{0.5\sqrt{0.69}}=-\frac{3}{\sqrt{69}}=-\frac{\sqrt{69}}{23}\approx -0.361[/tex]

Solution:
[tex]X=k[/tex] exactly when the first [tex]k-1[/tex] tosses give tails and the [tex]k[/tex]-th toss gives heads. By the independence of the tosses
[tex]P(X=k)=\left(\frac12\right)^{k-1}\times \frac12=\frac{1}{2^k},\qquad k=1,2,\ldots[/tex]
For [tex]|q|\lt 1[/tex] the geometric series [tex]\sum_{k=0}^{\infty}q^k=\frac{1}{1-q}[/tex] may be differentiated term by term inside its interval of convergence; this gives the first formula below. Multiplying it by [tex]q[/tex] and differentiating once more gives the second one:
[tex]\sum_{k=1}^{\infty}kq^{k-1}=\frac{1}{(1-q)^2},\qquad \sum_{k=1}^{\infty}k^2q^{k-1}=\frac{1+q}{(1-q)^3}[/tex]
With [tex]q=\frac12[/tex]:
[tex]E(X)=\sum_{k=1}^{\infty}\frac{k}{2^k}=\frac12\times \frac{1}{\left(\frac12\right)^2}=2,\qquad E(X^2)=\sum_{k=1}^{\infty}\frac{k^2}{2^k}=\frac12\times \frac{\frac32}{\left(\frac12\right)^3}=6[/tex]
Hence
[tex]D(X)=E(X^2)-(E(X))^2=6-4=2[/tex]
This agrees with the formula [tex]D(X)=\frac{1-p}{p^2}[/tex] for the geometric distribution with [tex]p=\frac12[/tex].

Solution:
Let [tex]X_1,\ldots,X_n[/tex] be the results, so that [tex]D(X_i)\lt 4[/tex], and let [tex]\bar X=\frac{X_1+\ldots+X_n}{n}[/tex]. By linearity of expectation and by independence
[tex]E(\bar X)=100,\qquad D(\bar X)=\frac{D(X_1)+\ldots+D(X_n)}{n^2}\lt\frac{4n}{n^2}=\frac{4}{n}[/tex]
Chebyshev's inequality with [tex]\varepsilon=4[/tex] gives
[tex]P(96\lt\bar X\lt 104)=P\left(|\bar X-100|\lt 4\right)\ge 1-\frac{D(\bar X)}{4^2}\gt 1-\frac{4}{16n}=1-\frac{1}{4n}[/tex]
It suffices that
[tex]1-\frac{1}{4n}\ge 0.98\iff \frac{1}{4n}\le 0.02\iff n\ge 12.5[/tex]
The smallest such integer is [tex]n=13[/tex]. For [tex]n=12[/tex] the bound, when the standard deviations are close to [tex]2[/tex], is close to [tex]1-\frac{1}{48}=\frac{47}{48}\approx 0.979\lt 0.98[/tex], so twelve measurements are not enough.

Solution:
First find [tex]C[/tex] from the normalization condition (the integral of the density over the plane equals [tex]1[/tex]):
[tex]\int_0^1\int_0^x (x+y)\,dy\,dx=\int_0^1\left(x^2+\frac{x^2}{2}\right)dx=\frac12\Rightarrow C=2[/tex]
Inside the triangle [tex]0\le y\le x\le 1[/tex] the event [tex]X+Y\lt 1[/tex] is the region [tex]0\le y\lt\frac12,\ y\le x\lt 1-y[/tex] (the lines [tex]y=x[/tex] and [tex]x+y=1[/tex] meet at [tex]\left(\frac12 , \frac12\right)[/tex]). Since [tex](x+y)^2[/tex] is an antiderivative of [tex]2(x+y)[/tex] with respect to [tex]x[/tex],
[tex]P(X+Y\lt 1)=\int_0^{1/2}\int_y^{1-y}2(x+y)\,dx\,dy=\int_0^{1/2}\left(1-4y^2\right)dy=\frac12-\frac16=\frac13[/tex]

Solution:
The parameter of an exponential distribution is the reciprocal of its mean, so the times are independent random variables [tex]X_1[/tex] and [tex]X_2[/tex] with densities [tex]e^{-t}[/tex] and [tex]\frac12e^{-t/2}[/tex] for [tex]t\ge 0[/tex]. The job takes [tex]T=\max(X_1,X_2)[/tex]. By independence, for [tex]t\ge 0[/tex]
[tex]P(T\le t)=P(X_1\le t)\,P(X_2\le t)=\left(1-e^{-t}\right)\left(1-e^{-t/2}\right)\Rightarrow P(T\gt t)=e^{-t}+e^{-t/2}-e^{-3t/2}[/tex]
For a nonnegative random variable [tex]E(T)=\int_0^{\infty}P(T\gt t)\,dt[/tex], hence
[tex]E(T)=\int_0^{\infty}\left(e^{-t}+e^{-t/2}-e^{-3t/2}\right)dt=1+2-\frac23=\frac73[/tex]
The expected time is [tex]\frac73[/tex] hours. (Equivalently, [tex]\max(X_1,X_2)=X_1+X_2-\min(X_1,X_2)[/tex], and [tex]\min(X_1,X_2)[/tex] is exponential with parameter [tex]1+\frac12=\frac32[/tex], i.e. with mean [tex]\frac23[/tex].)

Solution:
The order in which the balls come out is a uniformly random permutation of the [tex]5[/tex] balls, and the player gets exactly the balls standing before ball [tex]0[/tex] (ball [tex]0[/tex] itself adds nothing). For [tex]j=1,2,3,4[/tex] let [tex]I_j=1[/tex] if ball [tex]j[/tex] comes before ball [tex]0[/tex], and [tex]I_j=0[/tex] otherwise. Then the winnings are
[tex]S=1\times I_1+2I_2+3I_3+4I_4[/tex]
Swapping the positions of balls [tex]j[/tex] and [tex]0[/tex] is a bijection between the permutations in which [tex]j[/tex] is before [tex]0[/tex] and those in which it is after [tex]0[/tex], so
[tex]E(I_j)=P(I_j=1)=\frac12,\qquad j=1,2,3,4[/tex]
By linearity of expectation (which needs no independence)
[tex]E(S)=\frac{1+2+3+4}{2}=5[/tex]

Solution:
The box is chosen at random, so
[tex]P(Y=3)=\frac36=\frac12,\qquad P(Y=4)=\frac26=\frac13,\qquad P(Y=5)=\frac16[/tex]
Given [tex]Y=n[/tex], the number of hits [tex]X[/tex] is binomial with parameters [tex]n[/tex] and [tex]0.6[/tex], so [tex]E(X\mid Y=n)=0.6n[/tex]. Further,
[tex]E(Y)=\frac32+\frac43+\frac56=\frac{11}{3},\qquad E(Y^2)=\frac92+\frac{16}{3}+\frac{25}{6}=14[/tex]
By the total expectation formula
[tex]E(X)=\sum_n P(Y=n)\times 0.6n=0.6\,E(Y)=\frac{11}{5},\qquad E(XY)=\sum_n P(Y=n)\times n\times 0.6n=0.6\,E(Y^2)=\frac{42}{5}[/tex]
Hence
[tex]\operatorname{cov}(X,Y)=E(XY)-E(X)E(Y)=\frac{42}{5}-\frac{11}{5}\times \frac{11}{3}=\frac{126-121}{15}=\frac13[/tex]
(Equivalently, [tex]\operatorname{cov}(X,Y)=0.6\,D(Y)=0.6\times \frac59=\frac13[/tex].)

Solution:
Let [tex]N[/tex] be the number of arriving particles and [tex]M[/tex] the number of registered ones. Given [tex]N=n[/tex], the registrations are [tex]n[/tex] independent Bernoulli trials with success probability [tex]p[/tex], so for [tex]0\le k\le n[/tex]
[tex]P(M=k\mid N=n)=\binom{n}{k}p^k(1-p)^{n-k}[/tex]
and [tex]P(M=k\mid N=n)=0[/tex] for [tex]n\lt k[/tex]. By the law of total probability, for every integer [tex]k\ge 0[/tex]
[tex]P(M=k)=\sum_{n=k}^{\infty}\frac{\lambda^n}{n!}e^{-\lambda}\times \frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}[/tex]
Write [tex]\lambda^n=\lambda^k\lambda^{n-k}[/tex], substitute [tex]m=n-k[/tex] and use the series [tex]e^z=\sum_{m=0}^{\infty}\frac{z^m}{m!}[/tex], valid for every real [tex]z[/tex]:
[tex]P(M=k)=\frac{(\lambda p)^k}{k!}e^{-\lambda}\sum_{m=0}^{\infty}\frac{(\lambda(1-p))^m}{m!}=\frac{(\lambda p)^k}{k!}e^{-\lambda}e^{\lambda(1-p)}=\frac{(\lambda p)^k}{k!}e^{-\lambda p}[/tex]
These are exactly the probabilities of the Poisson distribution with parameter [tex]\lambda p[/tex], which proves the statement.