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X has a probability distribution:

6x(1-x), 0 < x < 1

0, elsewhere

Y has a probability distribution:

2y, 0 < y < 1

0, elsewhere

What is the CDF of $U$, where $U$ = $Y*X^2$?

I am trying to figure out how to start this. Any help would be appreciated.

Thanks in advance.

  • 0
    Please use [$\rm \LaTeX$](http://meta.math.stackexchange.com/q/5020/290189). What is $r$? I suppose you mean $f_Y(y) = 2y \forall 0 < y < 1$. – 2017-01-23
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    Hint: Compute the PDF of $(U,V)=(YX^2,X)$ using the Jacobian $dudv=x^2dxdy$, that is, $dxdy=dudv/v^2$ on the domain $0$U$ by marginalization. The answer is $$f_U(u)=6(1-\sqrt{u})^2\mathbf 1_{0 – 2017-01-23

1 Answers 1

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Given

The density functions for $X$ and $Y$ are \begin{align} f_X(x) &= \begin{cases} 6x(1-x) &\text{if } 0 < x < 1 \\ 0 &\text{elsewhere} \end{cases} \\ f_Y(y) &= \begin{cases} 2y &\text{if } 0 < y < 1 \\ 0 &\text{elsewhere.} \end{cases} \end{align}

Goal

To find the distribution function (a.k.a. CDF) of $U = X^2 Y$.

Approach

Use change of variables: we construct an auxiliary variable $V = V(X,Y)$ in order to get a bijective function $\phi:(0,1)\times(0,1)\to(0,1)\times(0,1)$ with $(u,v)=\phi(x,y)$, which gives $(x,y) = \phi^{-1}(u,v)$. We calculate the Jacobian $$\lvert J \rvert = \left\lvert\frac{\partial(u,v)}{\partial(x,y)}\right\rvert$$ and substitute it into $$f_{U,V}(u,v)=f_{X,Y}(x,y) \left\lvert\frac{\partial(x,y)}{\partial(u,v)}\right\rvert=f_{X,Y}(\phi^{-1}(u,v)) \lVert J \rVert^{-1}.$$

Find suitable boundaries for $u$ and $v$, so that we can integrate the joint distribution function $f_{U,V}(u,v)$ with respect to $v$ to obtain the marginal density function $f_U$, from which the CDF of $U$ can be obtained.

Details

For $(x,y) \in (0,1)\times(0,1)$, choose

\begin{cases} U&=X^2Y\\ V&=X. \end{cases}

So that we have the inverse transformation

\begin{cases} X&=V\\ Y&=\frac{U}{V^2} \end{cases}

Note that $0

\begin{align} \lvert J \rvert &= \left\lvert\frac{\partial(u,v)}{\partial(x,y)}\right\rvert \\ &= \left\lvert \begin{matrix} 2xy & x^2 \\ 1 & 0 \end{matrix}\right\rvert \\ &= -x^2 \end{align}

Since $X$ and $Y$ are independent variables, $f_{X,Y}(x,y) = f_X(x) f_Y(y)$. For $0 < \sqrt{u} < v < 1$,

\begin{align} f_{U,V}(u,v) &= f_X(v) f_Y \left(\frac{u}{v^2}\right) \lVert J \rVert^{-1} \\ &= 6v(1-v) \cdot \frac{2u}{v^2} \cdot \frac{1}{v^2} \\ &= 12u (v^{-3}-v^{-2}) \\ f_U(u) &= \int_{-\infty}^{+\infty} f_{U,V}(u,v)\,\mathrm{d}v \\ &= \int_{\sqrt{u}}^1 12u (v^{-3}-v^{-2})\,\mathrm{d}v \\ &= 12u \left[-\frac12 v^{-2} + v^{-1}\right]_{\sqrt{u}}^1 \\ &= 12u \left[\frac12 u^{-1}-\frac12 + 1 - u^{-1/2}\right] \\ &= 6(u - 2u^{1/2} + 1) \\ &= 6(1-u^{1/2})^2 \end{align}

For $0 < u < 1$, the CDF of $U$ is \begin{align} F_U(u) &= \int_{-\infty}^{+\infty} f_U(t)\,\mathrm{d}t \\ &= \int_0^u 6(t-2t^{1/2}+1)\,\mathrm{d}t \\ &= 6 \left( \frac{u^2}{2} - \frac43 u^{3/2} + u \right) \\ &= 3u^2 - 8u^{3/2} + 6u. \end{align}

To sum up, the CDF of $U$ is given by $$F_U(u)= \begin{cases} 0 &\text{if } u < 0 \\ 3u^2 - 8u^{3/2} + 6u &\text{if } 0 \le x \le 1 \\ 1 &\text{if } u > 1. \end{cases} $$

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    Nice approach. $ $ – 2017-01-23
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    Wow, I didn't even think of trying transformations using Jacobian. Thanks man. I was trying to integrate for both and using the joint of x and y. Way more work. – 2017-01-24