# Conditional Probability Uniform Bivariate Transformation Distribution

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

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

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 $$(V<.2|U=1.5)=$$

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

• 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. Jun 29 '20 at 7:52
• @StubbornAtom I added the conditional to my post. My process was correct, right? Sorry I know this is simple stuff, just been so long. Jun 29 '20 at 16:16
• Why in the joint PDF you have $0<V<4$? Shouldn't it be $0<V<2$ since $V=X$?
– user289381
Jun 29 '20 at 17:01
• I think you can find everything you need here youtube.com/watch?v=qUBlhsJpf1g
– user289381
Jun 29 '20 at 17:20
• @ping Oh, yes. It should be 0<V<2 and 0<V<U<V+2 right? Jun 29 '20 at 19:37

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