Examples

Sum of parabolic random variables

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

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

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

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

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

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

Final result is:

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

Discrete PMF formula for a sum

Verify the discrete formula for the PMF of a sum. (Apply the general formula for the PMF of $g(X,Y)$.)

Vandermonde’s identity from the binomial sum rule

Show that this “Vandermonde identity” holds for positive integers $n,m,\ell$:

$$\sum _{j+k=\ell }\left(\genfrac{}{}{0px}{}{n}{j}\right)\left(\genfrac{}{}{0px}{}{m}{k}\right)=\left(\genfrac{}{}{0px}{}{n+m}{\ell }\right)$$

Hint: The binomial sum rule is:

$$\text{“}\mathrm{Bin}(n,p)+\mathrm{Bin}(m,p)\sim \mathrm{Bin}(n+m,p)$$

Set $p=q=1/2$. Compute the PMF of the left side using convolution. Compute the PMF of the right side directly. Set these PMFs equal.

PDF of sums practice

Suppose $X$ is an RV with density:

$$f_X=\{\begin{array}{cc}2x& x\in [\mathrm{0,1}]\\ 0& \text{otherwise}\end{array}$$

Suppose $Y$ is uniform on $[\mathrm{0,1}]$ and independent of $X$.

Find the PDF of $X+Y$. Sketch the graph of this PDF.

Solution

(1) Write the CDF of $W=X+Y$ as a double integral:

$$F_W(w)=P[X+Y\le w]={\iint }_{x+y\le w}{f}_{X,Y}dxdy$$

The joint density on the unit square $x\in [\mathrm{0,1}],y\in [\mathrm{0,1}]$ is:

$$f_{X,Y}\gg \gg f_X\cdot f_Y\gg \gg 2x\cdot 1\gg \gg 2x$$

There is positive density in the region $x+y\le w$ only for $x\le w$ (otherwise $y<0$).

• When $w\in [\mathrm{0,1}]$, there is positive density in the region (only) when $y\le w-x$.
• When $w\in [\mathrm{1,2}]$, there is positive density in the region whenever $y\le 1$.

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(2) Evaluate $F_W(w)$ for $w\in [\mathrm{0,1}]$:

Here $x\in [0,w]$ and $w-x\le 1$, so $y\in [0,w-x]$.

$$\begin{array}{c}{F}_W(w)={\int }_0^w{\int }_0^{w-x}2xdydx\\ \\ \gg \gg {\int }_0^{w}2x(w-x)dx\gg \gg {[w{x}^2-\frac{2x^3}{3}]}_0^w\\ \\ \gg \gg w^3-\frac{2w^3}{3}\gg \gg \frac{w^3}{3}\end{array}$$

Differentiate:

$$f_W(w)=\frac{d}{dw}\frac{w^3}{3}\gg \gg w^2$$

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(3) Evaluate $F_W(w)$ for $w\in [\mathrm{1,2}]$:

Now $x$ ranges over $[\mathrm{0,1}]$. Split by whether the $y$-bound is $1$ or $w-x$:

• $x\in [0,w-1]$: $w-x\ge 1$, so $y\in [\mathrm{0,1}]$
• $x\in [w-1,1]$: $w-x<1$, so $y\in [0,w-x]$

$$\begin{array}{c}{F}_W(w)={\int }_0^{w-1}{\int }_0^{1}2xdydx+{\int }_{w-1}^1{\int }_0^{w-x}2xdydx\\ \\ \gg \gg {\int }_0^{w-1}2xdx+{\int }_{w-1}^{1}2x(w-x)dx\\ \\ \gg \gg (w-1{)}^2+{[w{x}^2-\frac{2x^3}{3}]}_{w-1}^1\\ \\ \gg \gg (w-1{)}^2+(w-\frac{2}{3})-(w(w-1{)}^2-\frac{2(w-1{)}^3}{3})\\ \\ \gg \gg -\frac{1}{3}{x}^3+w^2-\frac{1}{3}\end{array}$$

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(4) Differentiate for the final PDF:

$$f_W(w)=\frac{d}{dw}(-\frac{1}{3}{x}^3+w^2-\frac{1}{3})\gg \gg -w^2+2w$$

Therefore:

$$f_{X+Y}(w)=\{\begin{array}{cc}{w}^2& w\in [\mathrm{0,1}]\\ & \\ -w^2+2w& w\in [\mathrm{1,2}]\\ & \\ 0& \text{otherwise}\end{array}$$

Convolution practice

Suppose $X$ is an RV with density:

$$f_X=\{\begin{array}{cc}2x& x\in [\mathrm{0,1}]\\ 0& \text{otherwise}\end{array}$$

Suppose $Y$ is uniform on $[\mathrm{0,1}]$ and independent of $X$.

Find the PDF of $X+Y$. Sketch the graph of this PDF.

Solution

(1) Set up the convolution integral:

$$\begin{array}{c}{f}_{X+Y}(w)={\int }_{-\infty }^{+\infty }{f}_X(w-y)f_Y(y)dy\end{array}$$

Since $f_Y(y)=1$ on $[\mathrm{0,1}]$, this reduces to ${\int }_0^1{f}_X(w-y)dy$. The integrand $f_X(w-y)=2(w-y)$ is nonzero when $w-y\in [\mathrm{0,1}]$, i.e. $y\in [w-1,w]$. Intersecting with $[\mathrm{0,1}]$ gives the limits of integration.

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(2) Evaluate for $w\in [\mathrm{0,1}]$:

Integration region: $y\in [0,w]$.

$$\begin{array}{c}{f}_{X+Y}(w)={\int }_0^{w}2(w-y)dy\\ \\ \gg \gg {[2wy-y^2]}_0^w\gg \gg 2w^2-w^2\gg \gg w^2\end{array}$$

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(3) Evaluate for $w\in [\mathrm{1,2}]$:

Integration region: $y\in [w-1,1]$.

$$\begin{array}{c}{f}_{X+Y}(w)={\int }_{w-1}^{1}2(w-y)dy\\ \\ \gg \gg {[2wy-y^2]}_{w-1}^1\\ \\ \gg \gg (2w-1)-(2w(w-1)-(w-1{)}^2)\\ \\ \gg \gg 2w-1-(w^2-1)\gg \gg -w^2+2w\end{array}$$

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(4) Write the final piecewise PDF:

$$f_{X+Y}(w)=\{\begin{array}{cc}{w}^2& w\in [\mathrm{0,1}]\\ & \\ -w^2+2w& w\in [\mathrm{1,2}]\\ & \\ 0& \text{otherwise}\end{array}$$

Exp plus Exp equals Erlang - Without 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, and let $W=X+Y$.

Therefore:

$$f_{X,Y}(x,y)=\lambda e^{-\lambda x}\cdot \lambda e^{-\lambda y}={\lambda }^2{e}^{-\lambda (x+y)}x\ge 0,y\ge 0$$

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(1) Write the CDF as a double integral over the region $x\ge 0,y\ge 0,x+y\le w$:

For $w\ge 0$, the region is $x\in [0,w]$, $y\in [0,w-x]$.

$$F_W(w)={\int }_0^w{\int }_0^{w-x}{\lambda }^2{e}^{-\lambda (x+y)}dydx$$

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(2) Evaluate the inner integral:

$$\begin{array}{c}{\int }_0^{w-x}{\lambda }^2{e}^{-\lambda x}{e}^{-\lambda y}dy\gg \gg {\lambda }^2{e}^{-\lambda x}{[-\frac{1}{\lambda }{e}^{-\lambda y}]}_0^{w-x}\\ \\ \gg \gg \lambda e^{-\lambda x}(1-e^{-\lambda (w-x)})\\ \\ \gg \gg \lambda (e^{-\lambda x}-e^{-\lambda w})\end{array}$$

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(3) Evaluate the outer integral:

$$\begin{array}{c}{F}_W(w)=\lambda {\int }_0^w(e^{-\lambda x}-e^{-\lambda w})dx\\ \\ \gg \gg \lambda {[-\frac{1}{\lambda }{e}^{-\lambda x}-x{e}^{-\lambda w}]}_0^w\\ \\ \gg \gg \lambda [(-\frac{1}{\lambda }{e}^{-\lambda w}-w{e}^{-\lambda w})-(-\frac{1}{\lambda })]\\ \\ \gg \gg 1-e^{-\lambda w}-\lambda w{e}^{-\lambda w}\end{array}$$

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(4) Differentiate for the PDF:

$$\begin{array}{c}{f}_W(w)=\frac{d}{dw}(1-e^{-\lambda w}-\lambda w{e}^{-\lambda w})\\ \\ \gg \gg \lambda e^{-\lambda w}-\lambda e^{-\lambda w}+{\lambda }^{2}w{e}^{-\lambda w}\\ \\ \gg \gg {\lambda }^{2}w{e}^{-\lambda w}\end{array}$$

This is the $\mathrm{Erlang}(2,\lambda )$ density function:

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

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:

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

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:

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

Combining normals

Suppose $X\sim 𝒩(\mathrm{40,16})$, $Y\sim 𝒩(\mathrm{15,9})$. Find the probability that $X\ge 2Y$.

Solution

Define $W=X-2Y$. Using the formulas above, we see $W\sim 𝒩(\mathrm{10,52})$, or $W\sim \sqrt{52}Z+10$ for a standard normal $Z$. Then:

$$\begin{array}{c}P[X\ge 2Y]\gg \gg P[W\ge 0]\gg \gg P[Z\ge \frac{-10}{\sqrt{52}}]\\ \\ \gg \gg P[Z\le 1.39]\gg \gg \approx 0.918\end{array}$$