0
$\begingroup$

I'm reviewing probability theory from years ago and am a bit rusty. I'm not sure how to calculate the conditional probability for a uniform distribution after a bivariate transformation.

Suppose X and Y both follow a Uniform$(0,2)$ distribution. For the transformation $U=X+Y$ and $V=X$, I'll then have the following joint PDF: \begin{cases} \frac{1}{4}, & \text{if $0<V<U<4$}.\\ 0, & \text{otherwise}. \end{cases}

How would I calculate the conditional probability $P(V<.2|U=1.5)$? I'm likely missing something very intuitive. I've had someone say that U follows a $U(0,1.5)$ distribution and thus $P(V<.2) =\frac{1}{1.5} * .2$ but that doesn't seem right to me. I'd have to multiply by $\frac{1}{4}$ at some point, right?

Edit:

Conditional. on $f(V|U)$

First found the marginal of $U$ as follows:

\begin{cases} \frac{u}{4}, & \text{if $0<U<2$}\\ \frac{1}{2}-\frac{u-2}{4}, & \text{if $2<U<4$}\\ \ 0, & \text{otherwise} \end{cases}

Since $U=1.5$, the conditional $f(V|U=1.5)=\frac{f(U,V)}{f(U=1.5)}=\frac{\frac{1}{4}}{\frac{u}{4}}=\frac{1}{1.5}$

So the probability $(V<.2|U=1.5)=$

$$\int_{0}^{.2} \frac{1}{1.5} dv = .2*\frac{1}{1.5} - 0*\frac{1}{1.5}=.133$$

$\endgroup$
8
  • $\begingroup$ You are missing the independence of $X$ and $Y$. The distribution of $U$ is not uniform. Find the conditional density of $V$ given $U$ and hence the probability. $\endgroup$ Jun 29, 2020 at 7:52
  • $\begingroup$ @StubbornAtom I added the conditional to my post. My process was correct, right? Sorry I know this is simple stuff, just been so long. $\endgroup$
    – user627099
    Jun 29, 2020 at 16:16
  • $\begingroup$ Why in the joint PDF you have $0<V<4$? Shouldn't it be $0<V<2$ since $V=X$? $\endgroup$
    – user289381
    Jun 29, 2020 at 17:01
  • $\begingroup$ I think you can find everything you need here youtube.com/watch?v=qUBlhsJpf1g $\endgroup$
    – user289381
    Jun 29, 2020 at 17:20
  • $\begingroup$ @ping Oh, yes. It should be 0<V<2 and 0<V<U<V+2 right? $\endgroup$
    – user627099
    Jun 29, 2020 at 19:37

1 Answer 1

0
$\begingroup$

This follows my comment (using the approach shown in the Youtube video). Once you have the joint you can calculate the conditional.

Sorry for the typo. Of course, $g_2(X,Y)=X$, and not $Y$.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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