Quiz 8 Solution

Question 1

Let $X$ be a discrete random variable taking values in the set ${\mathcal{S}}_X=\{0,a\}$, where $a$ is a positive real number.  Suppose $Pr(X=a)=p$ for some $p\ne 0$.  Consider the event $X\ge a$.  Markov's inequality gives a bound $M$ on the probability of this event.  Let $T$ be the true probability of this event.  Find the ratio $M/T$.

(Hint: since Markov's inequality is an upper bound, we know that $T\le M$, so we must have $M/T\ge 1$.)

Solution:  Markov's inequality states, for a nonnegative random variable $X$ and a positive number $a$, that $Pr(X\ge a)\le \frac{E(X)}{a}$.   Let $M=E(X)/a$.  Here

$E(X)=\sum _{x\in {\mathcal{S}}_X}x{p}_X(x)=0\cdot (1-p)+a\cdot p=ap$,

and thus $M=\frac{ap}{a}=p$.  The true probability that $X\ge a$ is

$T=Pr(X\ge a)=\sum _{\begin{array}{c}x\in {\mathcal{S}}_X\\ x\ge a\end{array}}{p}_X(x)=p_X(a)=p$.

Thus we find $\frac{M}{T}=\frac{p}{p}=1$.  In other words, Markov's inequality is met with equality in this example.

Question 2

Fill in the blank.   On a particular streetcar route in Toronto, streetcars take an average of 60 minutes to travel from one end of the route to the other, with a standard deviation of 3 minutes.  Chebyshev's Inequality implies that at least _______% of streetcars take between 54 and 66 minutes to complete the route.

Solution:  Let $\sigma$ denote the standard deviation of a random variable $X$ having mean $m$.   Chebyshev's inequality states that

$Pr(|X-m|\ge k\sigma )\le \frac{1}{k^2}$.

Here, we observe that 54 is $k=2$ standard deviations below the mean of $m=60$ and $66$ is $k=2$ standard deviations above the mean of $m=60$.  Chebyshev's Inequality then gives

$Pr(|X-60|\ge 2\sigma )\le \frac{1}{4}$.  Looking at the complementary event, we get

$Pr(|X-60|<2\sigma )=Pr(54<X<66)=1-Pr(|X-60|\ge 2\sigma )\ge 1-\frac{1}{4}=\frac{3}{4}=75\mathrm{\%}$.  Thus the blank should be filled in with the number 75.

Question 3

The characteristic function ${\mathrm{\Phi }}_X(\omega )$ of a random variable $X$ is given as ${\mathrm{\Phi }}_X(\omega )=\frac{1}{4}(1+e^{j\omega }{)}^2$, where $j^2=-1$.  Find $E(X)$.

(Hint:  apply the moment theorem.)

Solution:

${\mathrm{\Phi }}_X^{\prime }(\omega )=\frac{1}{4}\cdot 2\cdot (1+e^{j\omega })\cdot e^{j\omega }\cdot j$.  Thus $E(X)=\frac{1}{j}{\mathrm{\Phi }}_X^{\prime }(0)=\frac{1}{j}\cdot \frac{1}{4}\cdot 2\cdot 2\cdot 1\cdot j=1$.

Question 4

Find $\text{VAR}(X)$, the variance of the random variable $X$ defined in the previous question.

Solution:

${\mathrm{\Phi }}_X^{″}(\omega )=\frac{j}{2}\cdot (e^{j\omega }\cdot j){\dot{e}}^{j\omega }+\frac{j}{2}(1+e^{j\omega })\cdot e^{j\omega }\cdot j=-\frac{1}{2}(1+2e^{j\omega })\cdot e^{j\omega }$.  Thus $E(X^2)=\frac{1}{j^2}{\mathrm{\Phi }}_X^{″}(0)=-{\mathrm{\Phi }}_X^{″}(0)=\frac{1}{2}\cdot (1+2)\cdot 1=\frac{3}{2}$.  It follows that $\text{VAR}(X)=E(X^2)-(E(X){)}^2=\frac{3}{2}-1^2=\frac{1}{2}$.