Question

We are given a stick that extends from 0 to . Its length, , is the realization of an exponential random variable , with mean . We break that stick at a point that is uniformly distributed over the interval .

Answer 1

Answer:- Given That:- We are given a stick that extends from 0 to x. Its length, x, is the realization of an exponential random variable X, with mean 1. We break that stick…

Answer 2

Answer:

The correct answer is -

1. $f_{X,Y}(x,y) = \left \{ {{\frac{e^{-x}}{x} ; 0 < y\leq x } \atop {0; elsewhere}} \right.$

2. Var[E(Y|X)] = $\frac{1}{4}$

Step-by-step explanation:

Complete Question -

1. Write down the joint PDF  $f_{X,Y} (x,y)$  of X and Y. For 0<y≤x.

2. Find Var(E[Y|X]).

Part - 1

Given that it is an exponential distribution -

Then,

$f(x) = e^{-x}$          ; x > 0

Break the stick at points that are Uniformly distributed throughout the interval.

$(Y|X=x) = U(0,x)$

$f(y|x) = \frac{1}{x}$  ; 0 < y ≤ x

Joint pdf $f_{X,Y} (x,y)$ of X and Y will be -

$f_{X,Y} (x,y) = f_X(x) f(y|x)$

$f_{X,Y}(x,y) = \left \{ {{\frac{e^{-x}}{x} ; 0 < y\leq x } \atop {0; elsewhere}} \right.$

Part - 2

To find E[Y|X] -

$E[Y|X] = \int\limits^x_0 {yf(x|y)} \, dy\\ \\E[Y|X] = \frac{1}{x}\times \frac{x^2}{2} = \frac{x}{2}$

Now we will find Var[E(Y|X)] -

$Var[E(Y|X)] = Var(\frac{x}{2} ) = \frac{1}{4} Var(x)$

Now,

$E(X^2) = \int\limits^{\infty}_0 {x^2e^{-x}} \, dx \\\\=\Gamma(3) = 2$(using gamma function)

Now,

To find Var(X) -

$Var(x) = E(X^2) - (E(X))^2\\\\Var(x) = 2 - 1 = 1$

Now putting the value in the equation we got above we get -

Var[E(Y|X)] = $\frac{1}{4}$

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