# Thread: Continuous Random Variable Question

1. ## Continuous Random Variable Question

Let U and V be independent and identically distributed uniformly on the interval [0,1]. Show that for 0<x<1: $P(x.

So far I have worked out the cdf, and hence pdf of $U^2$. I thought about $P(x but a) this doesn't give me the correct answer and b) I don't actually think P(V<U^2) can be calculated.

Could someone give me a hint as to where to go?
Thanks.

2. ## Re: Continuous Random Variable Question

why dont you think $P(U < V^2)$ can be calculated?

$P(U < V^2) = \int_{V=0}^{V=1} \int_{U=0}^{U=V^2} f(u,v) dU dV$

And similarly

$P(x < U < V^2) = \int_{V=x}^{V=1} \int_{U=x}^{U=V^2} f(u,v) dU dV$

(i think)

Assuming that integral is correct (check!), can you evaluate it?

3. ## Re: Continuous Random Variable Question

Hello !
Originally Posted by SpringFan25
why dont you think $P(U < V^2)$ can be calculated?

$P(U < V^2) = \int_{V=0}^{V=1} \int_{U=0}^{U=V^2} f(u,v) dU dV$

And similarly

$P(x < U < V^2) = \int_{V=x}^{V=1} \int_{U=x}^{U=V^2} f(u,v) dU dV$

(i think)

Assuming that integral is correct (check!), can you evaluate it?
There's something wrong with your capital letters. It should be small letters within the integrals !

The integral would then rather be :

$P(x < U < V^2) = \int_{v=x}^{v=\infty} \int_{u=x}^{u=v^2} f(u,v) \ du \ dv$

(since we have put f without any information, the indicator functions are still "in" it, so the boundaries go up to the infinity for v)

And f(u,v) is the joint pdf of U and V, which is the product of their pdf since they're independent.

4. ## Re: Continuous Random Variable Question

your right, but i rebel against lower case letters on thursdays

i think its reasonable to put the limits given in the question, its seems like matter of choice whether you define the pdf with an indicator function or over a limited range. (using an indicator function is an interesting idea actually, ive only ever sseen it done this way, but your way is neat).

5. ## Re: Continuous Random Variable Question

Shouldn't the limits for $v$ be $\sqrt x$ to 1?

I also prefer having the integration limits reflect the support of the pdf so that people don't make stupid mistakes when I tell them how to do a problem.

6. ## Re: Continuous Random Variable Question

yes i think you're right about the $\sqrt{x}$

7. ## Re: Continuous Random Variable Question

Thank you so much, came out as the right answer in the end!!