2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ "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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.