# Thread: Probability Bivariate Random Variable Problem

1. ## Probability Bivariate Random Variable Problem

Let U=min(X,Y) and V=max(X,Y), where X and Y are independent R(0,1) random variables.

1. Compute the cdf's (cumulative dist. function) F(u) and F(v) and hence derive the pdf's (probability density function) f(u) and f(v) of the random variables U and V respectively.

2. Compute the joint cdf F(u,v) ], 0<=u<=v<=1, of the bivariate random variable (U,V) and use it to derive the pdf f(u,v).

3. Are the random variables U and V independent? Why?

2. Originally Posted by maibs89
Let U=min(X,Y) and V=max(X,Y), where X and Y are independent R(0,1) random variables.

1. Compute the cdf's (cumulative dist. function) F(u) and F(v) and hence derive the pdf's (probability density function) f(u) and f(v) of the random variables U and V respectively.

2. Compute the joint cdf F(u,v) ], 0<=u<=v<=1, of the bivariate random variable (U,V) and use it to derive the pdf f(u,v).

3. Are the random variables U and V independent? Why?

I'll do a little bit to give you some ideas:

1. Consider $F(v) = {\color{red}\Pr(0 < V \leq v) = 1 - }\Pr(V > v)$.

If V > v then obviously X > v and Y > v.

So $F(v) = {\color{red}1 - }\Pr(X > v) \cdot \Pr(Y > v) = {\color{red}1 - }\left( \int_{v}^{1} 1 \, dx \right) \cdot \left( \int_{v}^{1} 1 \, dy \right) = ......$ if $0 \leq v \leq 1$ and zero otherwise.

Now recall that the pdf of V is given by $f(v) = \frac{dF}{dv}$ ......

3. Does f(u, v) = f(u) f(v) ?

Try the remaining bits again - post if you're still stuck after making a good attempt.

3. I got f(u)=min(xy) and f(v)=max(xy). Then later when I tried to compute the pdfs by differentiating I got one for both. I think that is not the answer.

I am even more confused on how to find the joint pdf. I know its f(u).f(v) but I don't know how to find part 1.

Help!!

4. Originally Posted by maibs89
I got f(u)=min(xy) and f(v)=max(xy). Then later when I tried to compute the pdfs by differentiating I got one for both. I think that is not the answer.

I am even more confused on how to find the joint pdf. I know its f(u).f(v) but I don't know how to find part 1.

Help!!
I have made a correction to my first reply. Please look at it. However, my original error is not the cause of your current difficulty .... I'm at a loss to understand how you could have got f(v) = 1. Please show your work (slightly revised to reflect my correction, of course).

5. I think I know how to get the cdfs, and hence pdfs:

F(v)= 1 - (1-v)^2 , 0<V<1
so I just differentiate to get f(v).

F(u)= u^2 , is it? Then the pdf f(u)=2u.

I need to know how to find the joint cdfs. Then I will know how to find independence.

Help!!!

6. Originally Posted by maibs89
I think I know how to get the cdfs, and hence pdfs:

F(v)= 1 - (1-v)^2 , 0<V<1
so I just differentiate to get f(v).

F(u)= u^2 , is it? Then the pdf f(u)=2u.

Mr F says: Correct on both counts.

I need to know how to find the joint cdfs. Then I will know how to find independence.

Help!!!
I'll give you some general direction:

$F(u, v) = \Pr(U \leq u ~ \text{and} ~ V \leq v | 0 \leq u \leq v \leq 1) = \Pr(U \leq u) \cdot \left( \, 1 - \Pr(V > v | v \geq u) \, \right)$.

$\Pr(V > v | v \geq u) = \frac{\Pr( V > v ~ \text{and} ~ V > u)}{\Pr(V > u)} = \frac{\Pr( V > v)}{\Pr(V > u)}$

and you can get Pr(V > u) from the pdf for V found in (a).

Then pdf = $f(u, v) = \frac{\partial^2 F(u, v)}{\partial u \partial v}$.

7. So far I understand.

How do I get Pr(V>V) and Pr(V>u)? What do yo mean by using the pdf in part a? Do I just
1-integrate for both? What do I integrate?

Help!!!

8. Originally Posted by maibs89
So far I understand.

How do I get Pr(V>V) and Pr(V>u)? What do yo mean by using the pdf in part a? Do I just
1-integrate for both? What do I integrate?

Help!!!
$\Pr(V > u) = \int_{u}^{1} f(v) \, dv$. f(v) is known from part (a).

9. I got the joint cdf as u^2 * (1-(v-1)^2/(u-1)^2)

Is this correct?

I am having difficulty integrating it to get the pdf and it looks complicated.

If it is the joint cdf, how would I integrate it? Thanks!

10. Originally Posted by maibs89
I got the joint cdf as u^2 * (1-(v-1)^2/(u-1)^2)

Is this correct?

I am having difficulty integrating it to get the pdf and it looks complicated.

If it is the joint cdf, how would I integrate it? Thanks!
The joint cdf looks OK.

You don't integrate it, you take partial derivatives of it. Recall: $f(u, v) = \frac{\partial^2 F(u, v)}{\partial u \partial v}$.

11. Originally Posted by mr fantastic
I'll do a little bit to give you some ideas:

1. Consider $F(v) = {\color{red}\Pr(0 < V \leq v) = 1 - }\Pr(V > v)$.

If V > v then obviously X > v and Y > v.

So $F(v) = {\color{red}1 - }\Pr(X > v) \cdot \Pr(Y > v) = {\color{red}1 - }\left( \int_{v}^{1} 1 \, dx \right) \cdot \left( \int_{v}^{1} 1 \, dy \right) = ......$ if $0 \leq v \leq 1$ and zero otherwise.

Now recall that the pdf of V is given by $f(v) = \frac{dF}{dv}$ ......

3. Does f(u, v) = f(u) f(v) ?

Try the remaining bits again - post if you're still stuck after making a good attempt.
I'm big enough to admit my mistakes. I made a mistake here. The pdf's are actually the other way around.

I only realised my mistake when answering the same question (more or less) asked by a different member at this thread: http://www.mathhelpforum.com/math-he...-question.html

After answering it, I thought it seemed familiar so I searched and found this thread. I was baffled that I'd given two mutually exclusive answers. Reading this thread again I saw the mistake in it immediately.

Sorry for any inconvenience. I did have the right ideas, just the wrong way around.