1

Let $X$ and $Y$ be independent with uniform distribution over $(0,a)$ and set $Z=X^2Y^2$. What is the joint density of $Z$ \begin{align} F_{Z}(t) &= \mathbb P(X^2Y^2<t) = \begin{cases} 0,& t<0\\ 1,& t>a^4 \end{cases} \end{align} Consider the case when $0<t<a^4$ $$F_{Z}(t) = \mathbb P(XY<\sqrt t)=\int\int_{xy<\sqrt t}f(x,y)dxdy=\int\int_{xy<\sqrt t}f_X(x)f_{Y}(y)dxdy =1/a^2 \int_{\sqrt t/a}^{a}dx\int_{\sqrt t/x}^{a}dy$$ Are the integration limits set correctly?

Sneach hcaens
  • 372

1 Answers1

1

No. You should integrate below the hyperbola $yx=\sqrt{t}$ or $y=\frac{\sqrt{t}}{a}$ at the picture:

enter image description here

So $$ F_{Z}(t) = \mathbb P(XY<\sqrt t)=1/a^2 \int_0^{\sqrt t/a}dx\int_0^a dy + 1/a^2 \int_{\sqrt t/a}^a dx\int_0^{\sqrt t/x} dy $$

Here is slightly modified picture with integration bounds for $y$ are drawn at each $x$.

enter image description here

NCh
  • 13,807
  • 4
  • 17
  • 33
  • So if $x$ from $0$ to $\sqrt t/a$ then $a$ y is higher then and i have to y from $0$ to $a$ but why when $x$ equals $\sqrt t/a$ $y$ equals $0$? $x = \sqrt t/a$ then $y = a$ I mean the integration limits in the second integral are from $a$ to $\sqrt t/x$ – Sneach hcaens Jan 27 '20 at 17:34
  • If $x=\sqrt{t}/a$ then any value of $0<y<a$ satisfy inequality $xy\leq \sqrt{t}$. Indeed, $$0<y<a \quad \text{then}\quad 0\cdot x < yx< ax=\sqrt{t} .$$ Do you see that $yx\leq \sqrt{t}$ $\iff$ $y\leq \frac{\sqrt{t}}{x}$ is the region below the hyperbola? To integrate over this region is to integrate separately: if $0<x<\sqrt{t}/a$ - left from dotted line, then any $0<y<a$ siuts us. For any $x$ right from dotted line, $y$ can vary from the bottom of the square (from $0$) to the hyperbola, i.e. to the $y=\sqrt{t}/x$. – NCh Jan 27 '20 at 18:12
  • Thank you very match, I was a little confused, but you really helped me. – Sneach hcaens Jan 27 '20 at 18:29
  • With pleasure. Just in case, I add a picture with integration bounds for $y$. Apologize for the lack of ability to draw. – NCh Jan 27 '20 at 18:38