W09 - Examples

Sums of random variables

Sum of parabolic random variables

Sum of parabolic random variables

Suppose $X$ is an RV with PDF given by:

$$f_X(x)=\{\begin{array}{ll}\frac{3}{4}(1-x^2)& x\in [-1,1]\\ 0& \text{otherwise}\end{array}$$

Let $Y$ be an independent copy of $X$. So $f_Y=f_X$, but $Y$ is independent of $X$.

Find the PDF of $X+Y$.

Solution

The graph of $f_X(w-x)$ matches the graph of $f_X(x)$ except (i) flipped in a vertical mirror, (ii) shifted by $w$ to the left.

When $w\in [-2,0]$, the integrand is nonzero only for $x\in [-1,w+1]$:

$$\begin{array}{rl}{f}_{X+Y}(w)& ={\left(\frac{3}{4}\right)}^2{\int }_{-1}^{w+1}(1-(w-x{)}^2\left)\right(1-x^2)dx\\ & =\frac{9}{16}\left(\frac{w^5}{30}-\frac{2w^3}{3}-\frac{4w^2}{3}+\frac{16}{15}\right)\end{array}$$

When $w\in [0,+2]$, the integrand is nonzero only for $x\in [w-1,+1]$:

$$\begin{array}{rl}{f}_{X+Y}(w)& ={\left(\frac{3}{4}\right)}^2{\int }_{w-1}^{+1}(1-(w-x{)}^2\left)\right(1-x^2)dx\\ & =\frac{9}{16}\left(-\frac{w^5}{30}+\frac{2w^3}{3}-\frac{4w^2}{3}+\frac{16}{15}\right)\end{array}$$

Final result is:

$$f_{X+Y}(w)=\{\begin{array}{ll}\frac{9}{16}\left(\frac{w^5}{30}-\frac{2w^3}{3}-\frac{4w^2}{3}+\frac{16}{15}\right)& w\in [-2,0]\\ & \\ \frac{9}{16}\left(-\frac{w^5}{30}+\frac{2w^3}{3}-\frac{4w^2}{3}+\frac{16}{15}\right)& w\in [0,2]\\ & \\ 0& \text{otherwise}\end{array}$$

Link to original

Exp plus Exp equals Erlang

Exp plus Exp equals Erlang - With Convolution

Let us verify this formula by direct calculation:

$$\text{“}\mathrm{Exp}(\lambda )+\mathrm{Exp}(\lambda )\sim \mathrm{Erlang}(2,\lambda )\text{”}$$

Solution

Let $X,Y\sim \mathrm{Exp}(\lambda )$ be independent RVs.

Therefore:

$$f_X=f_Y=\{\begin{array}{ll}\lambda e^{-\lambda x}& x\ge 0\\ 0& \text{otherwise}\end{array}$$

Now compute the convolution, assuming $w\ge 0$:

$$\begin{array}{c}{f}_{X+Y}(w)={\int }_{-\infty }^{+\infty }{f}_X(w-x)f_Y(x)dx\\ \\ \gg \gg {\int }_0^w{\lambda }^2{e}^{-\lambda (w-x)}{e}^{-\lambda x}dx\\ \\ \gg \gg {\lambda }^2{\int }_0^w{e}^{-\lambda w}dx\gg \gg {\lambda }^{2}w{e}^{-\lambda w}\end{array}$$

This is the Erlang PDF:

$$f_X(t)={\frac{{\lambda }^{\ell }}{(\ell -1)!}{t}^{\ell -1}{e}^{-\lambda t}|}_{\ell =2}$$ Link to original

Expectation for two random variables

Expectation of X squared plus Y from joint PMF chart

Expectation of X squared plus Y from joint PMF chart

Suppose the joint PMF of $X$ and $Y$ is given by this chart:

$Y\downarrow X\to$	$1$	$2$
$-1$	0.2	0.2
$0$	0.35	0.1
$1$	0.05	0.1

Define $W=X^2+Y$. Find the expectation $E[W]$.

Solution

First compute the values of $W$ for each pair $(X,Y)$ in the chart:

$Y\downarrow X\to$	$1$	$2$
$-1$	0	3
$0$	1	4
$1$	2	5

Now take the sum, weighted by probabilities:

$$\begin{array}{c}0\cdot (0.2)+3\cdot (0.2)+1\cdot (0.35)\\ +4\cdot (0.1)+2\cdot (0.05)+5\cdot (0.1)\end{array}\gg \gg 1.95=E[W]$$ Link to original

Expectation of Y two ways and Expectation of XY from joint density

Expectation of Y two ways and Expectation of XY from joint density

Suppose $X$ and $Y$ are random variables with the following joint density:

$$f_{X,Y}(x,y)=\{\begin{array}{ll}\frac{3}{16}x{y}^2& x,y\in [0,2]\\ 0& \text{ otherwise }\end{array}$$

(a) Compute $E[Y]$ using two methods.

(b) Compute $E[XY]$.

Solution

(a)

(1) Method One: via marginal PDF $f_Y(y)$:

$$f_Y(y)={\int }_0^2\frac{3}{16}x{y}^{2}dx\gg \gg \{\begin{array}{ll}\frac{3}{8}{y}^2& y\in [0,2]\\ 0& \text{otherwise}\end{array}$$

Then expectation:

$$E[Y]={\int }_0^{2}y{f}_Y(y)dy\gg \gg {\int }_0^2\frac{3}{8}{y}^{3}dy\gg \gg 3/2$$

(2) Method Two: directly, via two-variable formula:

$$E[Y]={\int }_0^2{\int }_0^{2}y\cdot \frac{3}{16}x{y}^{2}dydx\gg \gg {\int }_0^2\frac{3}{4}xdx\gg \gg 3/2$$

(b) Directly, via two-variable formula:

$$\begin{array}{c}E[XY]={\int }_0^2{\int }_0^{2}xy\cdot \frac{3}{16}x{y}^{2}dydx\\ \\ \gg \gg {\int }_0^2\frac{3}{4}{x}^{2}dx\gg \gg 2\end{array}$$Link to original

Covariance from PMF chart

Covariance from PMF chart

Suppose the joint PMF of $X$ and $Y$ is given by this chart:

$Y\downarrow X\to$	$1$	$2$
$-1$	0.2	0.2
$0$	0.35	0.1
$1$	0.05	0.1

Find $\mathrm{Cov}[X,Y]$.

Solution

We need $E[X]$ and $E[Y]$ and $E[XY]$.

$$E[X]=1\cdot (0.2+0.35+0.05)+2\cdot (0.2+0.1+0.1)\gg \gg 1.4$$ $$\begin{array}{c}E[Y]=-1\cdot (0.2+0.2)+0\cdot (0.35+0.1)+1\cdot (0.05+0.1)\\ \\ \gg \gg -0.25\end{array}$$ $$E[XY]=-1\cdot (0.2)-2\cdot (0.2)+0+1\cdot (0.05)+2\cdot (0.1)\gg \gg -0.35$$

Therefore:

$$\begin{array}{c}\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y]\\ \\ \gg \gg -0.35-(1.4)(-0.25)\gg \gg 0\end{array}$$ Link to original

Variance of sum of indicators

Variance of sum of indicators

An urn contains 3 red balls and 2 yellow balls.

Suppose 2 balls are drawn without replacement, and $X$ counts the number of red balls drawn.

Find $\mathrm{Var}[X]$.

Solution

Let $X_1$ indicate (one or zero) whether the first ball is red, and $X_2$ indicate whether the second ball is red, so $X=X_1+X_2$.

Then $X_1{X}_2$ indicates whether both drawn balls are red; so it is Bernoulli with success probability $\frac{3}{5}\frac{2}{4}=\frac{3}{10}$. Therefore $E[X_1{X}_2]=\frac{3}{10}$.

We also have $E[X_1]=E[X_2]=\frac{3}{5}$.

The variance sum rule gives:

$$\begin{array}{rl}\mathrm{Var}[X]& =\mathrm{Var}[X_1]+\mathrm{Var}[X_2]+2\mathrm{Cov}[X_1,X_2]\\ & \\ & \gg \gg E[X_1^2]-E[X_1{]}^2+E[X_2^2]-E[X_2{]}^2+2(E[X_1{X}_2]-E[X_1]E[X_2])\\ & \\ & \gg \gg \frac{3}{5}-{\left(\frac{3}{5}\right)}^2+\frac{3}{5}-{\left(\frac{3}{5}\right)}^2+2\left(\frac{3}{10}-\frac{3}{5}\cdot \frac{3}{5}\right)\gg \gg \frac{9}{25}\end{array}$$Link to original