0
$\begingroup$

Let $X$ and $Y$ be two absolutely continuous random variables on some probability space having joint distribution function $F$ and joint density $f$. Find the joint distribution and the joint density of the random variables $W=X^2$ and $Z=Y^2$.

I tried the following. I know that $$F(w,y)=P[W \leq w,Z \leq z]=P[X^2 \leq w,Y^2 \leq z]$$ Equivalently $$P[-\sqrt{w} \leq X \leq \sqrt{w},-\sqrt{z} \leq Y \leq \sqrt{z}]=\int_{-\sqrt{z}}^{\sqrt{z}}\int_{-\sqrt{w}}^{\sqrt{w}} f(x,y) \, dx \, dy$$ Then I tried the change of variable $x=\sqrt{w}$ but i can't see how to rescue the joint density and distribution of $W$ and $Y$

$\endgroup$
4
  • 1
    $\begingroup$ If $X$ and $Y$ are independent, then so are $X^2$ and $Y^2$. $\endgroup$
    – DirkGently
    Commented Oct 5, 2015 at 22:57
  • $\begingroup$ sorry they are not independent $\endgroup$ Commented Oct 5, 2015 at 23:03
  • $\begingroup$ @DirkGently : I think the way the problem is stated doesn't make sense if independence were assumed. ${}\qquad{}$ $\endgroup$ Commented Oct 6, 2015 at 1:17
  • $\begingroup$ @MichaelHardy, the original problem said explicitly that they were independent. The questioner has since corrected it. $\endgroup$
    – DirkGently
    Commented Oct 6, 2015 at 15:51

2 Answers 2

1
$\begingroup$

You want this: $$ f_{W,Z}(w,z) = \frac{\partial^2}{\partial w\,\partial z} \int_{-\sqrt{z}}^{\sqrt{z}}\int_{-\sqrt{w}}^{\sqrt{w}} f(x,y) \, dx \, dy. $$

Use the chain rule:

Write $u= \sqrt z$ so that $u^2 = z$ and $2u\,du = dz$ and then you have \begin{align} \frac{df}{dz} & = \frac{df}{du}\cdot\frac{du}{dz} = \left( \frac d {du} \int_{-u}^u \bullet\bullet\bullet \right) \cdot \frac d {dz} \sqrt z \\[10pt] & = \left( \left(\frac d {du} \int_0^u \bullet\bullet\bullet \right) - \left(\frac d {du} \int_0^{-u} \bullet\bullet\bullet\right) \right) \cdot \frac d {dz} \sqrt z \end{align}

You can handle $\displaystyle \frac d {du} \int_0^u \bullet\bullet\bullet$ via the fundamental theorem of calculus.

For $\displaystyle \frac d {du} \int_0^{-u} \bullet\bullet\bullet$ you can write $$ g = \int_0^v\bullet\bullet\bullet \text{ and } v=-u $$ and then find $\dfrac{dg}{dv}$ via the fundamental theorem, and $\frac{dv}{du}=-1$, and $$ \frac{dg}{du} = \frac{dg}{dv}\cdot\frac{dv}{du}. $$

$\endgroup$
0
$\begingroup$

$$\begin{align}\int_a^b\int_c^d f_{X,Y}(x,y)\operatorname d y \operatorname d x & =\int_a^b\int_c^d f_{X}(x) f_{Y\mid X=x}(y)\operatorname d y\operatorname d x \\[1ex] & = \int_a^b f_X(x)(F_{Y\mid X=x}(d)-F_{Y\mid X=x}(c))\operatorname d x \\[1ex] & = F_{X,Y}(b, d) - F_{X,Y}(a, d)- F_{X,Y}(b,c)+F_{X,Y}(a, c)\end{align}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .