Best Answer:
I'll give you first a long method and then a MUCH QUICKER one which is less error prone. The answer is the same with either method (of course): The covariance of U and V is 1/300.

The mean <U> of U is the sum <X>+<Y> of the means of X and Y. So, <U>=0.05+0.1=0.15.

Likewise, the mean of V is <V> = <Y>+<Z> = 0.1+0.15 = 0.25

The covariance of U and V is the mean of the product

(U-<U>) (V-<V>) = (X+Y-0.15)(Y+Z-0.25)

This is equal to <XY>+<XZ>-0.25<X>+<Y^2>+<YZ>-0.25<Y>-0.1...

For example, <XY>=<X><Y> = (0.05)(0.1) because the varaible X and Y are independent (their covariance is zero). Similarly, <XZ>=<X><Z> and <YZ>=<Y><Z>.

The only nontrivial term is <Y^2> which could be expressed in terms of the variance of Y or, directly, as the integral of y^2dy on the interval [0,0.2] divided by the length (0.2) of that interval. Namely: (.008/3)/(0.2) =4/300. If you plug this value and the aforementioned trivial values into the above expression and you would obtain the answer: 1/300.

However, there's an EASIER WAY. If you consider a fourth independent variable Y' distributed like Y, then the variables U and V'=Y'+Z are clearly independent. So the covariance of U and V' is zero. The expression of that zero covariance is exactly like the above, except that <YY'>=<Y><Y'>=<Y>^2 replaces <Y^2>.

Therefore, if we subtract from the above complicated expression for the covariance of U and V the zero expression which is the covariance of U and V' we obtain simply the expression <Y^2>-<Y>^2 (which is the variance of Y). We find, of course the same value for the covariance of U and V, but much more easilty:

cov(U,V) = <Y^2>-<Y>^2 = 1/75 - (1/10)^2 = 1/300 =0.0033333...

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