Vtyjkyuk

The problem asks to find the covariance. The joint density function is defined as,
$$
f(y_1,y_2)=
begincases
3y_1, & textfor $0 leq y_2 leq y_1 leq 1$, \
0, & textelsewhere. \
endcases
$$
My text defines
$$
operatornameCov(Y_1,Y_2)
= mathbbE(Y_1 Y_2) - mathbbE(Y_1)mathbbE(Y_2).
$$
I'm having problems finding $mathbbE(Y_2)$. My attempt is
begingather*
mathbbE(Y_2) = int_-infty^infty y_2 f_2(y_2) ,mathrmdy_2,
\
f_2(y_2) = int_-infty^infty f(y_1,y_2) ,mathrmdy_1
= int_0^1 3y_1 ,mathrmdy_1 = frac32,
\
mathbbE(Y_2) = int_0^1 y_2 frac32 ,mathrmdy_2
= frac34.
endgather*
But my book says $mathbbE(Y_2)= 3/8$. Where did I go wrong?

Pay attention when you evaluate the marginal distribution of $Y_2$. Observe that

$$0 leq y_2 leq y_1 leq 1.$$

Therefore, since you are integrating $y_1$ to get the marginal of $Y_2$, then the integration must be done on the set $y_2 leq y_1 leq 1$, not on $0 leq y_1 leq 1.$ In this case:

$$f_2(y_2)=int_-infty^infty f(y_1,y_2) dy_1
=int_y_2^13y_1dy_1=frac32(1-y_2^2),$$

valid for $0 leq y_2 leq 1.$

Then, the expected value of $Y_2$ is:
.
$$mathbbE(Y_2)=int_0^1 y_2 frac32(1-y_2^2)dy_2=dfrac38.$$

Same reasonings hold for the marginal of $Y_1$. In this case, you should consider $0 leq y_2 leq y_1.$

$$
f_Y_2(y_2) = int_y_2^1 3y_1, dy_1. text This should not be int_0^1.
$$
The main error here appears to be trying to solve math problems by proceeding algorithmically rather than by understanding. The educational system in effect encourages that, and professors have their heads buried deep in the sand about that issue.

To find $E(Y_2)$ you should use a double integral since the density function has two variables.
begineqnarray*
E(Y_2) &=& int_0^1 int_0^y_1 3 y_1 y_2 ,, dy_2 dy_1 \
E(Y_2) &=& int_0^1 frac 3 y_1 y_2^22 Bigg|_y_2=0^y_2=y_1 ,, dy_1 =
int_0^1 frac3y_1^32 ,, dy_1 \
E(Y_2) &=& frac3y_1^48 Big|_0^1 \
E(Y_2) &=& frac38
endeqnarray*

I hope this helps. Feel free to ask a follow up question.
Bob