Here's a problem I am stuck on:

Let $X$ and $Y$ be independent random variables such that $X$ is Bernoulli-distributed with $p=1/2$, and $Y$ is uniformly distributed on the interval $[0,1]$. Then:

  1. What is the CDF and PDF of $X+Y$ ?
  2. Does the PDF of $XY$ exist?
  3. What is the CDF of $XY$?

I tried finding the CDF of $X+Y$ by conditioning on $X$ as per this answer, but could not get any further. Can anyone show me what to do, or how to do this?

  • $\begingroup$ "but could not get any further." Please explain why. $\endgroup$ – Did May 22 '15 at 10:38
  • $\begingroup$ @Did I got $\sum_{x \in {0,1}} P(X+Y=k|X=i)(X=i)$, and rewrote the summand as $P(Y= k-i | X=i)P(X=i)$, which by independence is equal to $P(Y=k-i)P(X=i)$. Expanding the sum, I get $P(Y=k)P(X=0) + P(Y=k-1)P(X=1)$ $= (1/2)P(Y=k) + (1/2)P(Y=k-1) = (1/2)(P(Y=k)+P(Y=k-1))$ and I wasn't sure what to do next. $\endgroup$ – Newb May 22 '15 at 10:44
  • $\begingroup$ The conditioning argument works smoothly. Let $W=X+Y$. We calculate $\Pr(W\le w)$ if $1\lt w\le 2$. We have $W\le w$ if $X=0$ (prob. $1/2$) or $X=1$ and $Y\le w-1$ (prob. $(1/2)(w-1)$). Add. We get $w/2$. Now consider other ranges for $w$. $\endgroup$ – André Nicolas May 22 '15 at 10:48
  • $\begingroup$ Of course P(X+Y=k)=0 for every k. $\endgroup$ – Did May 22 '15 at 10:49
  • $\begingroup$ For the cdf you also need to deal with $w\lt 0$, $w\ge 2$, and $0\le w\le 1$. Separate treatment is in principle needed for $w=0$ and $w=1$. In the interval $(0,2)$ we get cdf $w/2$. Differentiate to get the density which is $1/2$. You differentiated $w/2$ incorrectly. The number $\frac{1}{2}$ is a constant, so the derivative of $\frac{1}{2}w$ is $\frac{1}{2}$ times the derivative of $w$, that is, $\frac{1}{2}$ times $1$. $\endgroup$ – André Nicolas May 22 '15 at 11:02

On 1)

Let $W:=X+Y$. Then:

$$F_{W}\left(w\right)=P\left(X+Y\leq w\mid X=0\right)P\left(X=0\right)+P\left(X+Y\leq w\mid X=1\right)P\left(X=1\right)=$$$$\frac{1}{2}F_{Y}\left(w\right)+\frac{1}{2}F_{Y}\left(w-1\right)$$

Here $F_{Y}$ is well known to you and knowing CDF $F_{W}$ you can find PDF $f_{W}$.

On 2)

$X=0\Rightarrow XY=0$ so that $P\left\{ XY=0\right\} \geq P\left\{ X=0\right\} \geq\frac{1}{2}$. Draw your conclusions about the existence of a PDF.

On 3)

Let $V:=XY$. Then:

$$F_{V}\left(v\right)=P\left(XY\leq v\mid X=0\right)P\left(X=0\right)+P\left(XY\leq v\mid X=1\right)P\left(X=1\right)=$$$$\frac{1}{2}P\left(0\leq v\right)+\frac{1}{2}F_{Y}\left(v\right)$$

Here $P\left(0\leq v\right)=0$ if $v<0$ and $P\left(0\leq v\right)=1$ otherwise.

  • $\begingroup$ Thanks -- I did some work in this vein with the help of Andre Nicolas. Could you please check what I did, as posted in the comments below my question? $\endgroup$ – Newb May 22 '15 at 12:16
  • $\begingroup$ In your comment concerning $X+Y$ you make mistakes. Just apply the formula in my answer to find the correct CDF and discern $4$ cases $w\leq0;0< w\leq1;1<w\leq 2;2<w$. Your result about CDF of $XY$ is correct. Your reasoning for no existence of a PDF is not the most adequate. The reasoning in my answer is better: If $P(V=c)>0$ for some constant $c$ then $V$ has no PDF. Here $V:=XY$. $\endgroup$ – drhab May 22 '15 at 12:31
  • $\begingroup$ If there is a PDF then the CDF must be continuous. $P(V=c)>0$ tells us that this is not the case, so there is no PDF. No that is not the correct CDF you mention. And where is the discernment in $4$ cases? $\endgroup$ – drhab May 22 '15 at 12:37
  • $\begingroup$ Not "wrong" but "narrow" I would say. E.g. there can be a PDF also if the CDF is not differentiable. So if possible then avoid it. $\endgroup$ – drhab May 22 '15 at 12:42
  • $\begingroup$ I see. Thanks for your help! If you'd like to briefly look over another problem I had, please view it here: math.stackexchange.com/questions/1293927/… $\endgroup$ – Newb May 22 '15 at 12:42

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