probability independent distribution

Number Cruncher 20 · #1 $Old$ November 17th, 2008, 07:37 AM

Hi,

Ive been given this question but i just can figure it out.

Suppose X,Y are Norm(1,1) and are independently and identically distributed.

Find the distribution of :

U = $(X-1)/(absolute value(Y-1))$

Any help would be really appreciated.

mr fantastic · #2 $Old$ November 18th, 2008, 02:22 AM

Quote:

Originally Posted by Number Cruncher 20 $View Post$

Hi,

Ive been given this question but i just can figure it out.

Suppose X,Y are Norm(1,1) and are independently and identically distributed.

Find the distribution of :

U = $(X-1)/(absolute value(Y-1))$

Any help would be really appreciated.

I'll get you started with a possible approach:

1. You know the joint pdf of X and Y, f(x,y).

2. Consider the cdf of U.

3. $U = \frac{X - 1}{|Y - 1|} = \frac{X-1}{Y-1}$ if $Y > 1$ and $\frac{X-1}{1 - Y}$ if $Y \leq 1$ so there are two cases to consider.

Case 1: $Y > 1$ .

Case 2: $Y \leq 1$ .

Therefore:

$\Pr(U < u) = \Pr\left( \frac{X - 1}{Y - 1} < u | Y > 1\right) \cdot \Pr(Y > 1)$ $+ \Pr\left( \frac{X - 1}{1 - Y} < u | Y \leq 1 \right) \cdot \Pr(Y \leq 1)$

$= \Pr(X - 1 < u[Y - 1] | Y > 1) \cdot \Pr(Y > 1) + \Pr(X - 1 < u[1 - Y] | Y \leq 1) \cdot \Pr(Y \leq 1)$

$= \Pr(X < u [Y - 1] + 1 | Y > 1) \cdot \Pr(Y > 1) + \Pr(X < u [1 - Y] + 1 | Y \leq 1) \cdot \Pr(Y \leq 1)$ .

Now integrate f(x, y) over the required regions and calculate each term.

4. The pdf of U is given by $g(u) = \frac{dF}{du}$ .