Find Average Time for Completion of Two Tasks with Joint MGF

[Tranquillity]

If the joint moment generating function of X and Y is

M X,Y (t1, t2) = 6 / (1-t1) * [(1/(2-t2)) -1/(3-(t1+t2))] and X,Y are the times AT WHICH the two successive tasks are completed, find the average time for completion of the two tasks.
Also find the moment generating function of the time needed for the second task and identify the distribution.For the first part I was thinking that to find the average time needed for the second task I should do E[Y] - E[X] if you can confirm and since task 1 starts at t=0 then E[X] will be the time required for the task 1 to be performed.

So the total average would be just E[Y]?

Concerning the second part I am not too sure how should I approach it!

Any help will be greatly appreciated. Thank you!

 

[Valkarie]

For the first part, you are correct that you can find the average time for completion of the two tasks by subtracting the expected value of X from the expected value of Y. To calculate the expected values, you will need to use the joint moment generating function, which is given by:M_X,Y(t1, t2) = 6 / (1-t1) * [(1/(2-t2)) - 1/(3-(t1+t2))]You can calculate E[X] and E[Y] by taking the partial derivatives of the MGF with respect to t1 and t2 respectively.E[X] = ∂M_X,Y/∂t1 = -6/(1-t1)^2 * [(1/(2-t2)) - 1/(3-(t1+t2))]E[Y] = ∂M_X,Y/∂t2 = 6/(1-t1) * [1/2(2-t2)^2 + 1/3(t1+t2)^2]So the average time for completion of the two tasks is E[Y] - E[X].For the second part, recall that the moment generating function of a random variable X is defined as M_X(t) = E[e^(tX)]. We can use this to calculate the moment generating function of the time needed for the second task, which we'll call T. We have:M_T(t) = E[e^(tT)] = E[e^(t(Y-X))] = E[e^(tY)e^(-tX)]The expectation is taken over the joint distribution of X and Y. By using the joint moment generating function, we can calculate the moment generating function of T as:M_T(t) = M_X,Y(t, -t) = 6/(1-t) * [(1/(2+t)) - 1/(3-(2t))]From this, we can identify the distribution of T as a mixture of the exponential and uniform distributions.