Random Variables and Distributions: Problems with Solutions

Solution:
The probabilities of all possible values add up to [tex]1[/tex], so
[tex]p_3=1-0.5-0.3=0.2[/tex]
By the definition of the expected value of a discrete random variable,
[tex]E(X)=x_1p_1+x_2p_2+x_3p_3=3\times 0.5+5\times 0.3+0.2\,x_3=3+0.2\,x_3[/tex]
Hence
[tex]3+0.2\,x_3=5\Rightarrow 0.2\,x_3=2\Rightarrow x_3=10[/tex]
The value [tex]10[/tex] differs from [tex]3[/tex] and [tex]5[/tex], so the three values are indeed distinct and the distribution of [tex]X[/tex] is
[tex]P(X=3)=0.5 , \quad P(X=5)=0.3 , \quad P(X=10)=0.2[/tex]

Solution:
Let [tex]X[/tex] be the exam score and [tex]Y[/tex] the module score, so [tex]D(X)=18[/tex], [tex]D(Y)=2[/tex] and [tex]\rho=\frac{1}{3}[/tex] (the means are not needed). By the definition of the correlation coefficient, the covariance is
[tex]\operatorname{cov}(X,Y)=\rho\sqrt{D(X)}\sqrt{D(Y)}=\frac{1}{3}\sqrt{18\times 2}=\frac{1}{3}\times 6=2[/tex]
The variance of a sum of two random variables is
[tex]D(X+Y)=D(X)+D(Y)+2\operatorname{cov}(X,Y)=18+2+4=24[/tex]

Solution:
Let [tex]a=E(X)[/tex] and [tex]\sigma=\sqrt{D(X)}\gt 0[/tex]. Chebyshev's inequality holds for every random variable with finite variance and every [tex]\varepsilon\gt 0[/tex]:
[tex]P(|X-a|\ge\varepsilon)\le\frac{D(X)}{\varepsilon^2}[/tex]
With [tex]\varepsilon=3\sigma[/tex] we get
[tex]P(|X-a|\ge 3\sigma)\le\frac{\sigma^2}{9\sigma^2}=\frac{1}{9}[/tex]
The event [tex]|X-a|\lt 3\sigma[/tex] is the complement of the event [tex]|X-a|\ge 3\sigma[/tex], hence
[tex]P(|X-a|\lt 3\sigma)\ge 1-\frac{1}{9}=\frac{8}{9}[/tex]
The bound holds for any distribution and cannot be improved in general: if [tex]X[/tex] takes the values [tex]a-3\sigma[/tex], [tex]a[/tex], [tex]a+3\sigma[/tex] with probabilities [tex]\frac{1}{18}[/tex], [tex]\frac{8}{9}[/tex], [tex]\frac{1}{18}[/tex], then its variance is [tex]2\times \frac{1}{18}\times 9\sigma^2=\sigma^2[/tex] and the probability is exactly [tex]\frac{8}{9}[/tex]. The value [tex]0.9973[/tex] (the three-sigma rule) is valid only for the normal distribution.

Solution:
By definition, the variance of a random variable [tex]Y[/tex] is
[tex]D(Y)=E\left[(Y-E(Y))^2\right][/tex]
For every random variable [tex]Y[/tex] with finite expected value we have [tex]E(cY)=cE(Y)[/tex], because the constant can be taken out of the sum (discrete case) or out of the integral (continuous case with density [tex]f[/tex]):
[tex]E(cY)=\sum_i c\,y_ip_i=c\sum_i y_ip_i,\qquad E(cY)=\int_{-\infty}^{\infty}c\,yf(y)\,dy=c\int_{-\infty}^{\infty}yf(y)\,dy[/tex]
Since [tex]D(X)[/tex] is finite, [tex]E(X)[/tex] exists, so [tex]E(cX)=cE(X)[/tex] and
[tex]cX-E(cX)=c\,(X-E(X))\Rightarrow (cX-E(cX))^2=c^2(X-E(X))^2[/tex]
Applying the same property to the random variable [tex](X-E(X))^2[/tex], whose expected value [tex]D(X)[/tex] is finite, we obtain
[tex]D(cX)=E\left[c^2(X-E(X))^2\right]=c^2E\left[(X-E(X))^2\right]=c^2D(X)[/tex]
which proves the statement. In particular, [tex]D(-X)=D(X)[/tex] and [tex]\sigma(cX)=|c|\,\sigma(X)[/tex].

Solution:
The number [tex]X[/tex] of passed tests has a binomial distribution with [tex]n=4[/tex], [tex]p=0.6[/tex], [tex]q=0.4[/tex]:
[tex]P(X=k)=\binom{4}{k}(0.6)^k(0.4)^{4-k}[/tex]
This gives
[tex]P(X=0)=\frac{16}{625} , \quad P(X=1)=\frac{96}{625} , \quad P(X=2)=\frac{216}{625}[/tex]
[tex]P(X=3)=\frac{216}{625} , \quad P(X=4)=\frac{81}{625}[/tex]
The maximum is attained at two values, [tex]k=2[/tex] and [tex]k=3[/tex]. In general,
[tex]\frac{P(X=k+1)}{P(X=k)}=\frac{(n-k)p}{(k+1)q}[/tex]
and for [tex]k=2[/tex] this ratio equals [tex]\frac{2\times 0.6}{3\times 0.4}=1[/tex]. This happens because [tex](n+1)p=3[/tex] is an integer; then both [tex](n+1)p-1=2[/tex] and [tex](n+1)p=3[/tex] are most probable values. Note that the expected value [tex]np=2.4[/tex] is not even a possible value of [tex]X[/tex].

Solution:
All [tex]\binom{6}{3}=20[/tex] choices of [tex]3[/tex] devices are equally likely. There are only [tex]2[/tex] non-standard devices, so at least one chosen device is standard and [tex]X\in\{1, 2, 3\}[/tex]. The variable [tex]X[/tex] has a hypergeometric distribution:
[tex]P(X=k)=\frac{\binom{4}{k}\binom{2}{3-k}}{\binom{6}{3}}[/tex]
hence
[tex]P(X=1)=\frac{4\times 1}{20}=\frac{1}{5} , \quad P(X=2)=\frac{6\times 2}{20}=\frac{3}{5} , \quad P(X=3)=\frac{4\times 1}{20}=\frac{1}{5}[/tex]
(the sum is [tex]1[/tex]). Then
[tex]E(X)=1\times \frac{1}{5}+2\times \frac{3}{5}+3\times \frac{1}{5}=2[/tex]
[tex]E(X^2)=1\times \frac{1}{5}+4\times \frac{3}{5}+9\times \frac{1}{5}=\frac{22}{5}[/tex]
[tex]D(X)=E(X^2)-(E(X))^2=\frac{22}{5}-4=\frac{2}{5}[/tex]
Check with the formula for the variance of the hypergeometric distribution ([tex]N=6[/tex], [tex]M=4[/tex], [tex]n=3[/tex]):
[tex]D(X)=n\frac{M}{N}\left(1-\frac{M}{N}\right)\frac{N-n}{N-1}=3\times \frac{2}{3}\times \frac{1}{3}\times \frac{3}{5}=\frac{2}{5}[/tex]

Solution:
First find [tex]c[/tex] from the normalization condition [tex]\int_{-\infty}^{\infty}f(x)\,dx=1[/tex]. The integrand is even, so
[tex]\int_{-8}^{8}c\,(8-|x|)\,dx=2c\int_0^8(8-x)\,dx=2c\times 32=64c=1\Rightarrow c=\frac{1}{64}[/tex]
Now [tex]F(x)=\int_{-\infty}^{x}f(t)\,dt[/tex]. For [tex]x\lt -8[/tex] the density is zero, so [tex]F(x)=0[/tex]. For [tex]-8\le x\le 0[/tex] we have [tex]|t|=-t[/tex], hence
[tex]F(x)=\frac{1}{64}\int_{-8}^{x}(8+t)\,dt=\frac{(x+8)^2}{128}[/tex]
In particular [tex]F(0)=\frac{1}{2}[/tex], as expected from the symmetry of the density. For [tex]0\lt x\le 8[/tex]:
[tex]F(x)=\frac{1}{2}+\frac{1}{64}\int_0^x(8-t)\,dt=\frac{1}{2}+\frac{x}{8}-\frac{x^2}{128}=1-\frac{(8-x)^2}{128}[/tex]
For [tex]x\gt 8[/tex] the whole probability mass lies to the left of [tex]x[/tex], so [tex]F(x)=1[/tex]. Check: [tex]F(-8)=0[/tex], [tex]F(8)=1[/tex], and [tex]F[/tex] is continuous and non-decreasing. Therefore
[tex]F(x)=\begin{cases}0, & x\lt -8\\ \frac{(x+8)^2}{128}, & -8\le x\le 0\\ 1-\frac{(8-x)^2}{128}, & 0\lt x\le 8\\ 1, & x\gt 8\end{cases}[/tex]