Question

(Statistics):

Answer 1

it's so hard to type this question out, so i attached a picture of the question instead, my question is that i need help on parts b) and probably c) unless the answer to b) helps me answer c)

Answer 2

So for b:
Recall that in general, if \(X,Y\) are jointly distributed with pdf \(f_{X,Y}(x,y)\), then the marginal pdf of \(X\) is
\[ f_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\, dy\]

In your case then, since \(Y_1, Y_2\) are jointly distributed, the marginal pdf of \(Y_1\)is\[ f_{Y_1}(y_1)=\int_{-\infty}^{\infty}f_{Y_1,Y_2}(y_1,y_2)\, dy_2\]
Now, you found the result in a) for the joint pdf \(f_{Y_1,Y_2}(y_1,y_2)=f_{X_1,X_2}(y_1-y_2,y_2)\) so just substitute that into b) to get

\[ f_{Y_1}(y_1)=\int_{-\infty}^{\infty}f_{X_1,X_2}(y_1-y_2,y_2)\, dy_2\]
--------------------------------------------
For c) Recall in the general case that if \(X,Y\) are independent, then \(f_{X,Y}(x,y)=f_X(x)f_Y(y)\). So in your case, you have that:

\(f_{X_1,X_2}(y_1-y_2,y_2)=f_{X_1}(y_1-y_2)f_{X_2}(y_2)\)

So just plug in that result into the answer for b) to get:

\[ f_{Y_1}(y_1)=\int_{-\infty}^{\infty}f_{X_1}(y_1-y_2)f_{X_2}(y_2)\, dy_2\]