Answer :
(a) The joint probability distribution of the two points will be 1/x, 0<x<1, 0<y<1
(b) The expectation of the product of the two points will be 1/6
(c) The marginal distribution of the point your friend picks will be (-1/2) lny^2
We have, X=(0,1)
Y | X=X=(0, X)
a) Let f(x,y) be the joint probability distribution of the two points will be:
f(X)=1
f(Y|X)=1/x
[tex]$\begin{aligned}& f(x, y)=f(X) f(Y \mid X=x)=1 / x, 0 < x < 1,0 < y < x\end{aligned}$[/tex]
b) Let E(XY) be the expectation of the product of the two points.
[tex]$\begin{aligned}E(XY)=& \int_0^1 \int_0^x \frac{x y}{x} d y d x=\frac{1}{6} \quad \text { (Decimal : } 0.16667 \ldots \text { ) }\end{aligned}$[/tex]
c) Let f(y) be the marginal distribution of the point your friend picks.
So, the value of f(y) will be:
[tex]$\begin{aligned}& \mathrm{f}(\mathrm{y})= \\& \int_y^1 \frac{1}{x} d x=-\frac{1}{2} \ln \left(y^2\right)\end{aligned}$[/tex]
d)for 0<y<1
[tex]$\begin{aligned}& f(y) > =0 \\& \int_0^1-\frac{1}{2} \ln \left(y^2\right) d y=1\end{aligned}$[/tex]
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