Home page for accesible maths 8 Bivariate Transformations 8.1 One-to-one Bivariate Transformations 8.3 Links Between Standard Distributions

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8.2 Use of Dummy Variables

Often we are interested in not two new variables, $S$ and $T$ , but in just one, $S$ say. To obtain the pdf of $S$ alone we have to create a dummy variable $T$ , obtain the joint pdf of $S$ and $T$ , then integrate to get the marginal distribution of $S$ .

1.

Define a new variable $T$ which makes a one-to-one bivariate transformation between $(X,Y)$ and $(S,T)$ . The choice of $T$ is essentially arbitrary and can be made for convenience. Sometimes some trial and error is required.
2.

Find the joint pdf $f_{ST}(s,t)$ of $S$ and $T$ .
3.

Find the marginal pdf of $S$ :

$\displaystyle{\color[rgb]{0.76,0.01,0}f_{S}(s)=\int f_{ST}(s,t)\,\mathrm{d}t}$

taking care with the range of integration.

Example 8.2.1.

Example 8.1.1 showed that if $(X,Y)$ are independent $\operatorname{\mathsf{N}}(0,1)$ then $S=X+Y$ and $T=X-Y$ are both marginally $\operatorname{\mathsf{N}}(0,2)$ (also $S$ and $T$ are independent, but that is irrelevant to what follows). Ignoring $T$ we see that $S=X+Y\sim\operatorname{\mathsf{N}}(0,2)$ .

Example 8.2.2.

If $(X,Y)$ have joint pdf

\displaystyle f_{XY}(x,y)={1\over x^{2}y^{2}}

for $x>1$ and $y>1$ , find the pdf of $S=XY$ .

With $S=XY$ put $T={\color[rgb]{0.76,0.01,0}X}$ . Clearly $S>1$ and $T>1$ . However $Y>1$ , so $~{}{\color[rgb]{0.76,0.01,0}S/T>1}~{}$ , so $~{}{\color[rgb]{0.76,0.01,0}S>T}~{}$ . So the joint range is ${\color[rgb]{0.76,0.01,0}S>T>1.}$

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The inverse is

$X={\color[rgb]{0.76,0.01,0}T}$ ,
$Y={\color[rgb]{0.76,0.01,0}\frac{S}{X}=\frac{S}{T}}$ .

so that

\displaystyle\begin{bmatrix}{\partial s\over\partial x}&{\partial s\over% \partial y}\\ {\partial t\over\partial x}&{\partial t\over\partial y}\end{bmatrix}={\color[% rgb]{0.76,0.01,0}\begin{bmatrix}{\color[rgb]{0.76,0.01,0}y}&{\color[rgb]{% 0.76,0.01,0}x}\\ {\color[rgb]{0.76,0.01,0}1}&{\color[rgb]{0.76,0.01,0}0}\end{bmatrix}}

and $|\det J|={\color[rgb]{0.76,0.01,0}1/x}$ .

The joint pdf of $(S,T)$ is

	$\displaystyle f_{ST}(s,t)$	$\displaystyle={\color[rgb]{0.76,0.01,0}f_{XY}(x,y)1/x\left.\right\|_{x=t,y=s/t}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{x^{3}y^{2}}\left.\right\|_{x=t,% y=s/t}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{s^{2}t}}$

for $1<t<s<\infty$ . The marginal pdf of $S$ is

	$\displaystyle f_{S}(s)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{t=1}^{s}\frac{1}{s^{2}t}\,\mathrm% {d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\left[\frac{\log(t)}{s^{2}}\right]_{t=1% }^{s}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\log(s)}{s^{2}}}$

for $1<s<\infty$ . To check this integrates to $1$ :

	$\displaystyle\int_{1}^{\infty}\frac{\log(s)}{s^{2}}\,\mathrm{d}s$	$\displaystyle=\left[-\frac{\log(s)}{s}\right]_{1}^{\infty}+\int_{1}^{\infty}% \frac{1}{s^{2}}\,\mathrm{d}s$
		$\displaystyle=\left[-\frac{1}{s}\right]_{1}^{\infty}=1.$

Convolution

A transformation of general interest is $S=X+Y$ . We use the dummy variable method to obtain the pdf of $S$ . We make the transformation

$S=X+Y$ ,
$T=X$ ,

with $T$ as the dummy variable. It follows that the inverse transformation is

$X=T$ ,
$Y=S-T$ ,

\displaystyle\left|\det{\frac{\partial(x,y)}{\partial(s,t)}}\right|=\left|\det% \begin{bmatrix}0&1\\ 1&-1\end{bmatrix}\right|=\mid-1\mid=1.

Thus

\displaystyle f_{ST}(s,t)=f_{XY}(t,s-t),

so the marginal pdf of $S=X+Y$ is

\displaystyle f_{S}(s)=\int_{-\infty}^{\infty}f_{XY}(t,s-t)\,\mathrm{d}t.

This formula is known as the convolution formula. It is finding the probability of $S=X+Y$ by summing the probabilities, over all possible $t$ , for the pairs $(t,s-t)$ in $(X,Y)$ .

Example 8.2.3.

Let $X\sim{\operatorname{\mathsf{Unif}}}(0,1)$ and $Y\sim{\operatorname{\mathsf{Unif}}}(0,1)$ , find the density of $S=X+Y$ using the convolution formula.

Solution. For $0\leq x\leq 1$ , $0\leq y\leq 1$ ,

\displaystyle f_{XY}(x,y)=1.

The range restrictions on $x$ and $y$ imply that $f_{X,Y}(t,s-t)$ is only non-zero when $0\leq t\leq 1$ and $0\leq s-t\leq 1$ . The latter implies $s-1\leq t\leq s$ . $S$ can only be between $0$ and $2$ . When $~{}s\leq 1~{}$ the restrictions simplify to $~{}{\color[rgb]{0.76,0.01,0}0\leq t\leq s}~{}$ , whereas when $~{}s>1~{}$ they simplify to $~{}{\color[rgb]{0.76,0.01,0}s-1\leq t\leq 1.}~{}$ Denote the lower and upper bounds for $t$ by $a_{s}$ and $b_{s}$ for now.

	$\displaystyle f_{S}(s)$	$\displaystyle=\int_{-\infty}^{\infty}f_{XY}(t,s-t)\,\mathrm{d}t$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{a_{s}}^{b_{s}}1\,\mathrm{d}t}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}b_{s}-a_{s}}$
		$\displaystyle=\left\{\begin{array}[]{ll}0&\quad s\leq 0\\ {\color[rgb]{0.76,0.01,0}s}&\quad 0<s\leq 1\\ {\color[rgb]{0.76,0.01,0}2-s}&\quad 1<s\leq 2\\ 0&\quad s>2\end{array}\right.$

The density is a triangle.

Example 8.2.4.

Exam2016 Let $X\sim\operatorname{\mathsf{Beta}}(\alpha,1)$ ( $\alpha\neq 1$ ) and $Y\sim{\operatorname{\mathsf{Unif}}}(0,1)$ be independent of each other. The point $(X,Y)$ defines the top-right corner of a rectangle whose bottom-left corner is at the origin, as shown in the figure below. Let $\mathcal{A}$ be this rectangle, let $A$ be its area and let $V=X$ .

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(a)

Without performing any calculation, write down the conditional distribution of $A$ given $X=x$ .

Solution. $A|X=x~{}\sim{\color[rgb]{0.76,0.01,0}{\operatorname{\mathsf{Unif}}}(0,x)}$ (since ${\color[rgb]{0.76,0.01,0}A=XY}$ and ${\color[rgb]{0.76,0.01,0}Y\sim{\operatorname{\mathsf{Unif}}}(0,1).}$ )
(b)

Derive the joint density of $A$ and $V$ , $f_{A,V}(a,v)$ ; be sure to state clearly the joint range of $A$ and $V$ .

Solution. The range (which most students got wrong) is $~{}{\color[rgb]{0.76,0.01,0}0<a<v<1}~{}$ since $0<y<1$ but $y=a/v$ . The inverse map is $X=V$ , $Y=A/V$ , so

$\displaystyle J=\frac{\partial(x,y)}{\partial(a,v)}={\color[rgb]{0.76,0.01,0}% \begin{bmatrix}{\color[rgb]{0.76,0.01,0}0}&{\color[rgb]{0.76,0.01,0}1}\\ {\color[rgb]{0.76,0.01,0}1/v}&{\color[rgb]{0.76,0.01,0}-a/v^{2}}\end{bmatrix}}$

Then $|\det J|={\color[rgb]{0.76,0.01,0}1/v}$ , so with the range as defined above,

$\displaystyle f_{A,V}(a,v)$ $\displaystyle={\color[rgb]{0.76,0.01,0}f_{X,Y}(x,y)\frac{1}{v}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\alpha x^{\alpha-1}\frac{1}{v}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\alpha v^{\alpha-2}.}$
(c)

Show that the marginal density of $A$ is $f_{A}(a)=\frac{\alpha}{\alpha-1}(1-a^{\alpha-1})$ for $(0<a<1)$ and $0$ elsewhere.

Solution. For $0<a<1$ ,

$\displaystyle f_{A}(a)={\color[rgb]{0.76,0.01,0}\int_{a}^{1}\alpha v^{\alpha-2% }\,\mathrm{d}v=\frac{\alpha}{\alpha-1}\left[v^{\alpha-1}\right]_{a}^{1},}$

which simplifies to the required expression. $A$ must be between $0$ and $1$ since it is the product of two rvs which have these bounds.
(d)

Find the conditional density of $V$ given $A=a$ .

Solution.

$\displaystyle f_{V|A}(v|a)={\color[rgb]{0.76,0.01,0}\frac{\alpha v^{\alpha-2}}% {\frac{\alpha}{\alpha-1}\left(1-a^{\alpha-1}\right)}=\frac{\alpha-1}{1-a^{% \alpha-1}}v^{\alpha-2}}$

for $a<v<1$ .
(e)

A new co-ordinate pair $(X^{\prime},Y^{\prime})$ is chosen, where $X^{\prime}\sim{\operatorname{\mathsf{Unif}}}(0,1)$ and $Y^{\prime}\sim{\operatorname{\mathsf{Unif}}}(0,1)$ are independent of each other and of $X$ and $Y$ . Show that the probability that $(X^{\prime},Y^{\prime})$ is in the (random) rectangle $\mathcal{A}$ is $p_{\alpha}=\frac{\alpha}{2(\alpha+1)}$ .

Solution. The point $(X^{\prime},Y^{\prime})$ has a uniform distribution in the unit square. So, conditional on knowing $\mathcal{A}$ , $\operatorname{\mathsf{P}}\left({(X^{\prime},Y^{\prime})\in\mathcal{A}|\mathcal% {A}}\right)=~{}{\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({(X^{% \prime},Y^{\prime})\in\mathcal{A}|A}\right)=A}~{}$ , the area of ${\color[rgb]{0.76,0.01,0}\mathcal{A}}$ . Since for any event $B$ , $\operatorname{\mathsf{P}}\left({B,A\in(a,a+\,\mathrm{d}a]}\right)=% \operatorname{\mathsf{P}}\left({B|a}\right)f_{A}(a)\,\mathrm{d}a$ , exactly as for any joint probability, we can marginalise by integrating out $a$ :

$\displaystyle\operatorname{\mathsf{P}}\left({(X^{\prime},Y^{\prime})\in% \mathcal{A}}\right)$ $\displaystyle={\color[rgb]{0.76,0.01,0}\int_{a=0}^{1}\operatorname{\mathsf{P}}% \left({(X^{\prime},Y^{\prime})\in\mathcal{A}|a}\right)f_{A}(a)\,\mathrm{d}a}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\alpha}{\alpha-1}\int_{0}^{1}% \left(a-a^{\alpha}\right)\,\mathrm{d}a}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\alpha}{\alpha-1}\left[\frac{1}{2% }a^{2}-\frac{a^{\alpha+1}}{\alpha+1}\right]_{0}^{1},}$

which simplifies to the required expression.
(f)

Note that $\lim_{\alpha\rightarrow 1}p_{\alpha}=1/4$ . Provide a derivation of this particular value using simple probability calculations (i.e. without resorting to the calculations similar to those in Parts (b)-(e)).

Solution. When $\alpha=1$ , $X$ and $X^{\prime}$ have the same distribution, so $\operatorname{\mathsf{P}}\left({X^{\prime}<X}\right)={\color[rgb]{0.76,0.01,0}% 1/2}$ . Similarly $\operatorname{\mathsf{P}}\left({Y^{\prime}<Y}\right)={\color[rgb]{0.76,0.01,0}% 1/2}$ thus

$\displaystyle\operatorname{\mathsf{P}}\left({(X^{\prime},Y^{\prime})\in% \mathcal{A}}\right)=\operatorname{\mathsf{P}}\left({X^{\prime}<X,Y^{\prime}<Y}% \right)={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X^{\prime}<X}% \right)\operatorname{\mathsf{P}}\left({Y^{\prime}<Y}\right)}$

by independence . So $\operatorname{\mathsf{P}}\left({(X^{\prime},Y^{\prime})\in\mathcal{A}}\right)=% {\color[rgb]{0.76,0.01,0}1/4.}$

	$\displaystyle f_{A,V}(a,v)$	$\displaystyle={\color[rgb]{0.76,0.01,0}f_{X,Y}(x,y)\frac{1}{v}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\alpha x^{\alpha-1}\frac{1}{v}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\alpha v^{\alpha-2}.}$

	$\displaystyle\operatorname{\mathsf{P}}\left({(X^{\prime},Y^{\prime})\in% \mathcal{A}}\right)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\int_{a=0}^{1}\operatorname{\mathsf{P}}% \left({(X^{\prime},Y^{\prime})\in\mathcal{A}\|a}\right)f_{A}(a)\,\mathrm{d}a}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\alpha}{\alpha-1}\int_{0}^{1}% \left(a-a^{\alpha}\right)\,\mathrm{d}a}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{\alpha}{\alpha-1}\left[\frac{1}{2% }a^{2}-\frac{a^{\alpha+1}}{\alpha+1}\right]_{0}^{1},}$