joint pdf

[anon_econ]

The probability density of x is given by
f(x)= 1+x for -1<x<=0
       1-x for 0<x<1
       0 otherwise

U=X and V=X^2. Find the joint pdf of U and V.

[SHIKHA]

@vasudha
were u able to find the solution????...if yes..den let me know..

[Amit Goyal]

Joint CDF is computed as follows:
F(u, v)
= Pr(U < u, V < v)
= Pr(X < u, X^2 < v)  

Case 1: u < -1
F(u, v) = 0

Case 2: v < 0
F(u, v) = 0

Case 3: u ≥ 1, v ≥ 1
F(u, v) = 1

Case 4: -1 ≤ u < 1, v ≥ 1
F(u, v) = Pr(X < u)

Case 5:  u ≥ 1, 0 ≤ v < 1,
F(u, v) = Pr(X^2 < v) = Pr(-v^{.5} < X < v^{.5})

Case 6: -1 ≤ u < 1, 0 ≤ v < 1
F(u, v) = Pr(X < u, X^2 < v)
= 0 if u ≤ -v^{.5}
= Pr(-v^{.5} < X < min{v^{.5}, u}) if u > -v^{.5}

Now just use f(x) to find the above. And then compute the joint pdf by differentiating cdf wrt u and v.

[AJ]

Will we have further sub-cases in 4th and 6th case...??

like for 4th:

Joint CDF is
(1/2 + u + u^2/2) when -1<u<=0
(1/2 + u - u^2/2) when 0<u<1

Is this right..??

And m getting 3 subcases for 6th case...???

[AJ]

If above is right.. my joint PDF is coming out to be:

Pr(u, v)= 0,                   when u < -1
             0,                    when v < 0
             0,                   when u ≥ 1, v ≥ 1
             1+u,                 when -1<u<=0, v ≥ 1
             1-u,                 when 0<u<1  , v ≥ 1
             2(v)^{0.5} - v,  when  u ≥ 1, 0 ≤ v < 1
             

these are for cases till 5.. are these correct..???
please someone reply..

[Amit Goyal]

AJ,

For cases 1 to 5, PDF f(u, v) which is the derivative of CDF F(u, v) with respect to u AND v is:
f(u, v) = 0. (Just think about it and you will know your mistake)

[AJ]

Do we have to take double derivative... firstly with respect to either u or v... and then wrt the other one..???

If this is so, then even for 6th case I am getting f(u,v)=0

Is my CDF correct for case 4..??

[anon_econ]

Thanks sir. I've understood this. Will try to get the pdf when i get some time after tomorrow's exam.

[Amit Goyal]

Actually, You can also compute joint PDF directly.

And think why it is the PDF:

g(u, v) = 1 + u, if -1 < u ≤ 0, and v = u^2
          = 1 - u, if 0 < u ≤ 1, and v = u^2
          = 0, otherwise.

[anon_econ]

Ok sir. This seems so obvious but I couldn't come up with it myself!
Anyway thanks :)