2
$\begingroup$

I am preparing for an exam. This is a previous exam paper question.

Given $U \sim N(0,1)$ and $W\sim b(1,\frac{1}{2})$. Is $UW$ a discrete random variable? They are also independent.

I tried to do the following. Set $Z=UW$:

\begin{align} P(Z\leq z) &= P(UW \leq z)\\ &= P(UW \leq z, W=1) + P(UW \leq z, W=0)\\ &= P(U \leq z, W=1) + 0\\ &= P(U \leq z) \cdot P(W=1)\\ &= \frac{1}{2} \cdot P(U \leq z) \end{align}

And by this point I wasn't sure how to continue, and am pretty sure I messed something up. This is not homework, so please don't be afraid to reveal the answer, I already know what it is I just can't get there.

  • 1
    With $b(1, 1/2)$ you mean Bernoulli with probability of success 1/2?2017-01-05
  • 1
    If $P(W=0)=P(W=1)=\frac12$ and $(U,W)$ is independent then $UW$ is neither discrete nor continuous since its distribution has a part of each type. Using the CDF to try to understand what is going on is, here as in many other cases on this site, perhaps the most ineffective way there is.2017-01-05
  • 0
    @JimmyR. Yes. Was my notation unclear? I apologise. What should I have written instead?2017-01-05
  • 0
    @k4kuz0 Yes, a little bit, $W\sim Ber(p=\frac12)$ would me more clear. Actually it is obvious from your calculation that $W$ is Bernoulli, but I saw this before reading your whole post. So, it's ok as it is, I think2017-01-05

1 Answers 1

1

The equality $P(UW\le z, W=0)=0$ is true only if $z<0$. If $z\ge 0$ you have that $$P(UW\le z, W=0)=P(0\le z)=1$$ Hence for $z\ge 0$ you have that \begin{align}P(Z\le z)&=\dots\\[0.2cm]&=P(UW\le z, W=1)+P(UW\le z,W=0)\\[0.2cm]&=P(U\le z)P(W=1)+1\cdot P(W=0)\\[0.2cm]&=\frac12P(U\le z)+\frac12\end{align} In sum and using the notation $P(U\le z)=\Phi(z)$ since $U$ is standard normal, we have $$F_Z(z)=\begin{cases}\frac12\Phi(z), &z<0\\\frac12\Phi(z)+\frac12, &z\ge0\end{cases}$$ Hence, there is a mass in $z=0$ which means that the random variable $Z$ is not atomless.

  • 0
    So, to clarify: The answer is no, it is not a discrete random variable nor a continuous random variable.2017-01-05