(a) Verify that f(x, y) = { 4xy 0 . is a joint density function. (b) If X and Y a

step 1 To begin the problem, let's write down the hypothesis. So we have the density, F of X, comma, y,

(a) Verify that f(x, y) = { 4xy 0 . is a...

(a) Verify that $ f(x, y) = \left\{ \begin{array}{ll} 4xy & \mbox{if $ 0 \le x \le 1, 0 \le y \le 1 $}\\ 0 & \mbox{otherwise} \end{array} \right.$ is a joint density function. (b) If $ X $ and $ Y $ are random variables whose joint density function is the function $ f $ in part (a), find (i) $ P (X \ge \frac{1}{2}) $ (ii) $ P (X \ge \frac{1}{2}, Y \le \frac{1}{2}) $ (c) Find the expected values of $ X $ and $ Y $.

(a) Verify that

$ f(x, y) = \left\{
\begin{array}{ll}
4xy & \mbox{if $ 0 \le x \le 1, 0 \le y \le 1 $}\\
0 & \mbox{otherwise}
\end{array} \right.$

is a joint density function.
(b) If $ X $ and $ Y $ are random variables whose joint density function is the function $ f $ in part (a), find
(i) $ P (X \ge \frac{1}{2}) $ (ii) $ P (X \ge \frac{1}{2}, Y \le \frac{1}{2}) $
(c) Find the expected values of $ X $ and $ Y $.

Transcript

00:01 To begin the problem, let's write down the hypothesis.

00:05 So we have the density, f of x, comma, y, is equal to 4xy, when 0 less than equal to x less than equal to 1, then 0 less than equal to y, less than equal to 1, and 0 otherwise.

00:26 So the first observation is, if of x, comma, y is greater than equal to 0, which is clear, because x times y is greater than equal to 0 when x and y lies in this range and then we have to check the integration 0 to 1 4xy then d y followed by t x is equal to 1 so let's evaluate this integration and show that this is indeed is equal to 1 so this becomes 4 0 to 1 and the integration of y is y squared divided by 2.

01:08 So x times y squared divided by 2, 0 to 1 followed by d x.

01:14 So 4 integration 0 to 1, x divided by 2 followed by d x.

01:23 So this becomes 2, and the integration of x is simple x squared divided by 2, 0 to 1, and this becomes 1.

01:32 So this verifies the second axiom for the density function so if of x comma y is indeed and it should be a a density of x and y so from this let's find out the density of x so the density of x is going to be if capital x small x so you have to integrate over the other variable so 0 to 1 4xy d y so this becomes 4x followed by y squared divided by 2 0 to 1 which is simply 2x similarly so the density of y is is going to be 2 times y by the similar computation why we need this because we have to compute the mean in part c so for part b one so we have to compute probability x is greater than equal to half if x is greater than equal to half we have already computed the density for x so this is going to be half to one two x d x so x square because the integration of two x is x square half to one so one minus one over 4 which is 3 over 4 next part b 2 we have to compute probability of x is greater than equal to half and y is less than equal to half so this is going to be integration so x is greater than equal to half so half to 1 and the y is less than equal to half so this should be 0 to half 4 xy and d y d x okay so then half to 1 and this will the integration of y is simply y squared divided by 2 so x y square divided by 2 i have 4 then 0 to half t x i can cancel that 2 so this becomes half to 1 2x if i put the upper limit this is simply 1 over 4 right so then followed by dx i can cancel the 2 again so this becomes 2 so let's take the half outside so this becomes half integration let's go back so half to 1 and then x d x so now this is x squared divided by 2 half to 1 so 1 over 2 so 1 minus if i take 1 over 2 square so excuse me that should be half because upper limit is half and the lower limit is going to be so 2 square is 4 so 1 over 8 so this can be written as 1 over 4 1 minus 1 over 4 so 1 over 4 so 1 over 4 times 3 over 4 is equal to 3 over 16...

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