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Let $U \sim e(1)$ and $V$ a discrete random variable independent of $U$ such that $p_V(v)=1/2$ if $v \in \{-1,1\}$ and $p_V(v)=0$ otherwise.

Problem: Let $W=UV$. Find the distribution function of $W$ $\forall w\in \mathbb{R}$.

My try: \begin{align} P(W\leq w)&=P(UV\leq w)=P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1)\\ &= P(U\leq w)\frac{1}{2}+P(U\geq -w)\frac{1}{2}\\ &=P(U\leq w)\frac{1}{2}+(1-P(U\leq-w)\frac{1}{2}\\ &=\left(\int_0^we^{-u} \, du\right)\frac{1}{2}+\left(1-\int_0^{-w} e^{-u} \, du \right) \frac{1}{2}\\ &=(1-e^{-w})\frac{1}{2}+\frac{1}{2}e^w = F_W(w) \end{align}

The correct answer in my book is the following $$F_W(w)=\frac{1}{2}(1-e^{-w})+\frac{1}{2}, w>0$$ and $$F_W(w)=\frac{1}{2}e^{w}, w\leq 0$$

Can anyone tell me what i am doing wrong?

Michael Hardy
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Mathe
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    If you write $F_w(w)$ instead of $F_W(w),$ then what's the point of the subscript? The notation $F_W(w)$ means $\Pr(W\le w),$ and that would be incomprehensible if one failed to distinguish between $W$ and $w.$ The notations $F_X(3),$ $F_Y(3),$ and $F_W(3)$ mean different things, and that's why that subscript is there. – Michael Hardy Jan 01 '18 at 22:05
  • Your first mistake is to replace $$P(U\leq w)$$ by $$\int_0^we^{-u} , du$$ These coincide if $w\ge0$ but not if $w<0$. – Did Jan 01 '18 at 22:16

2 Answers2

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You wrote: \begin{align} & P(W\leq w) = P(UV\leq w) \\[10pt] = {} &P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1). \end{align}

What is $P(U\le w\mid V=1)$?

It is $\begin{cases} 1-e^{-w} & \text{if } w\ge0, \\ 0 & \text{if } w<0. \end{cases}$

What is $P(-U\le w\mid V=-1)$?

It is $\begin{cases} e^w & \text{if } w\le0, \\ 1 & \text{if } w>0. \end{cases}$

You need to consider the cases $w\ge0$ and $w<0$ separately.

If you find the density function, you will be able to write it more simply by using the notation $|w|.$

Michael Hardy
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Hint: An exponentially distributed random variable has positive support.

Raskolnikov
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