joint distribution

ques 6.23
Couldn't understand last step i.e how we get gamma dist.
Ques6.24
And how do we get that variables are not independent.One reason can be becoz covariance are given so we are using formula for it.What if examiner tries to confuse by giving covariances.

 

6.23: Look at the pdf of Gamma given in the tables. Plug in alpha = 3 and see if you are getting the pdf as mentioned in the solution

6.24:
Recall:
(1) Independence implies covariance (and correlation) = 0
(2) Covariance (and correlation) = 0 does not imply independence at all times.
However, Covariance (and correlation) <> 0 does imply not independent

 

well how we got the answer when we put lamda=mu in the result we got after integration.plz explain
Thanks for replying.

 

You can't plug in lambda = mu in the result which you have obtained assuming they are not equal (see the first line of the solution).

Rather plug in lambda = mu in the first line before you start integrating i.e.

\[f_{Z}(z) = \intop_0^z \lambda e^{-\lambda x} \cdot \lambda e^{-\lambda (z - x)} dx\]
or, \[f_{Z}(z) = \intop_0^z \lambda^2 e^{-\lambda z} dx\]
or, \[f_{Z}(z) = \lambda^2 z e^{-\lambda z}\]

This is what you are looking for.

 

thanks