Question

Q: Assistance in understanding and solving this example from Probability and Statistical (Conditional Distributions) with the steps of the solution to better understand, thanks.

**Please give the step by steps with details to completely see how the solution came about.

1) Suppose X and Y both take values in [0,1] with joint probability density f(x,y) = 4xy.

a) Find f_x(x) and f_y(y), the marginal probability density functions.

b) Are the two random variables independent? Why or why not?

c) Compute the means and variances of X and Y.

d) Find P(X $\leq$ Y).

Answer 1

1)

a)

$f(x) = \int f(x,y)dy = \int 4xy dy = 2x$

similarly

$f(y) = \int f(x,y)dx= \int 4xy dx = 2y$

b)

Yes , two variables are independent as

f(x,y) = f(x) f(y)

c)

$E(X) = \int x f(x) dx =$

$\int _0^1x\cdot \:2xdx=\frac{2}{3}\quad \left(\mathrm{Decimal:\quad }\:0.66667\dots \right)$

Var(X) = E(X^2) - (E(X))^2

$E(X^2) = \int _0^1x^2\cdot \:2xdx=\frac{1}{2}\quad \left(\mathrm{Decimal:\quad }\:0.5\right)$

hence

Var(X) = 1/2 - (2/3)^2

= 0.05555

E(Y) = E(X) = 2/3

Var(Y) = Var(X) = 0.05555    {X and Y are symmetric}

d)

$P(X \leq Y)= \int _0^1\int _0^x4xydydx=\frac{1}{2}\quad \left(\mathrm{Decimal:\quad }\:0.5\right)$

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