Quiz 9 Solution

Let $X$ and $Y$ be independent exponentially-distributed random variables of unit mean.  Thus $X$ and $Y$ have marginal probability density functions

$f_X(x)=\{\begin{array}{ll}\exp (-x),& x\ge 0;\\ 0,& \text{otherwise},\end{array}$ and $f_Y(y)=\{\begin{array}{ll}\exp (-y),& y\ge 0;\\ 0,& \text{otherwise}.\end{array}$

Question 1.

Determine the correlation coefficient ${\rho }_{X,Y}$ of $X$ and $Y$.

Answer: Since $X$ and $Y$ are independent, it follows that $X$ and $Y$ are uncorrelated.  As such, ${\rho }_{X,Y}=0$.

Question 2.

Determine $Pr(max(X,Y)\le \ln (2))$.

Answer:  We have $max(X,Y)\le \ln (2)$ if and only if $X\le \ln (2)$ and $Y\le \ln (2)$.  This is the product form event $(X,Y)\in (-\mathrm{\infty },\ln (2)]\times (-\mathrm{\infty },\ln (2)]$ shown in the diagram.  We can calculate the probability of this event by integrating the joint pdf.  Since $X$ and $Y$ are independent, the joint pdf is the product of the marginal pdfs.

We get $Pr(max(X,Y)\le \ln 2)={\int }_{-\mathrm{\infty }}^{\ln 2}{\int }_{-\mathrm{\infty }}^{\ln 2}{f}_{X,Y}(x,y)\text{d}y\text{d}x$

$={\int }_{-\mathrm{\infty }}^{\ln 2}{\int }_{-\mathrm{\infty }}^{\ln 2}{f}_X(x)f_Y(y)\text{d}y\text{d}x$  (since $X$ and $Y$ are independent)

$={\int }_0^{\ln 2}{\int }_0^{\ln 2}{f}_X(x)f_Y(y)\text{d}y\text{d}x$  (since the marginal pdfs are zero for negative values of their argument)

$={\int }_0^{\ln 2}{f}_X(x)\text{d}x\cdot {\int }_0^{\ln 2}{f}_Y(y)\text{d}y$

$={({\int }_0^{\ln 2}\exp (-x)\text{d}x)}^2$

$={\left({-\exp (-x)|}_0^{\ln 2}\right)}^2$

$={(1-\exp (-\ln (2)))}^2$

$={(1-\exp (\ln (1/2)))}^2$

$={(1-\frac{1}{2})}^2$

$=\frac{1}{4}$.

Question 3.

Determine $Pr(min(X,Y)\le \ln (2))$.

Answer:  We have $min(X,Y)\le \ln (2)$ if $X\le \ln (2)$ or $Y\le \ln (2)$.  This corresponds to the shaded region in the diagram below.

It is easier to integrate over the complementary region.  Thus we find

$Pr(min(X,Y)\le \ln (2))=1-{\int }_{\ln 2}^{\mathrm{\infty }}{\int }_{\ln 2}^{\mathrm{\infty }}\exp (-x)\exp (-y)\text{d}x\text{d}y$

$=1-{\left({-\exp (-x)|}_{\ln 2}^{\mathrm{\infty }}\right)}^2$

$=1-(\exp (-\ln (2){)}^2$

$=1-\frac{1}{4}$

$=\frac{3}{4}$.

Question 4.

Determine $Pr(X+Y\le \ln (2))$.

Answer:  We have $X+Y\le \ln 2$ in the shaded region shown below.

Since the joint pdf is zero outside the first quadrant, we integrate only over the triangular region in the first quadrant to obtain:

$Pr(X+Y\le \ln (2))={\int }_{x=0}^{\ln 2}{\int }_{y=0}^{-x+\ln 2}\exp (-x)\exp (-y)\text{d}y\text{d}x$

$={\int }_0^{\ln 2}\exp (-x)\left({-\exp (-y)|}_0^{-x+\ln 2}\right)\text{d}x$

$={\int }_0^{\ln 2}\exp (-x)(1-\exp (x-\ln 2))\text{d}x$

$={\int }_0^{\ln 2}\exp (-x)\text{d}x-{\int }_0^{\ln 2}\exp (-\ln 2)\text{d}x$

$=(1-\exp (-\ln 2))-\ln (2)\exp (-\ln 2)$

$=1-\frac{1}{2}-\frac{1}{2}\ln (2)$

$=\frac{1}{2}(1-\ln (2))$

$\approx 0.1534264$

Question 5.

Determine $E((X-Y{)}^2$.

Answer:  Expanding the square we get

$E((X-Y{)}^2)=E(X^2-2XY+Y^2)$

$=E(X^2)-2E(XY)+E(Y^2)$

$=E(X^2)-2E(X)E(Y)+E(Y^2)$ (since $X$ and $Y$ are independent, they are uncorrelated)

$=2E(X^2)-2(E(X){)}^2$  (since $X$ and $Y$ have the same distribution)

$=2\text{VAR}(X)$

$=2$ (since $X$ has a variance of $1$).

(Why does $X$ have a variance of $1$?  An exponential random variable with parameter $\lambda$, having a pdf $f(x)=\lambda \exp (-\lambda x)$ when $x\ge 0$, has a mean of $\frac{1}{\lambda }$ and a variance of $\frac{1}{{\lambda }^2}$.  Here $\lambda =1$.)