# conditioning problem - homework due very soon

• Mar 3rd 2010, 09:02 PM
paulrb
conditioning problem - homework due very soon
Suppose X|Y ~ poisson (mean = y) and Y ~ Uniform [0, 5]

compute Pr [X >= 1]

I think I did it, but need confirmation it's right

E[X] = 2.5

Pr [X >= 1 ] = 1 - e^(-2.5)

this seems overly simple, but is it correct?
• Mar 4th 2010, 09:07 PM
CaptainBlack
Quote:

Originally Posted by paulrb
Suppose X|Y ~ poisson (mean = y) and Y ~ Uniform [0, 5]

compute Pr [X >= 1]

I think I did it, but need confirmation it's right

E[X] = 2.5

Pr [X >= 1 ] = 1 - e^(-2.5)

this seems overly simple, but is it correct?

$\displaystyle Pr(X\ge 1)=1-Pr(X=0)=1-\int_{\text{support}(y)} Pr(X=0|y)p(y)\;dy=$ $\displaystyle 1-\int_{y=0}^5 \frac{Pr(X=0|y)}{5}\;dy$ $\displaystyle =1-\frac{1}{5}\int_{y=0}^5 e^{-y} \;dy$

CB
• Mar 4th 2010, 10:48 PM
matheagle
But $\displaystyle f_Y(y)=1/5$ on 0<y<5

$\displaystyle Y\sim U(0,5)$ not $\displaystyle U(0,1)$
• Mar 5th 2010, 01:54 AM
CaptainBlack
Quote:

Originally Posted by matheagle
But $\displaystyle f_Y(y)=1/5$ on 0<y<5

$\displaystyle Y\sim U(0,5)$ not $\displaystyle U(0,1)$

Oppss ... working on auto-pilot there.

CB
• Mar 8th 2010, 03:07 AM
chialin4
Quote:

Originally Posted by CaptainBlack
$\displaystyle Pr(X\ge 1)=1-Pr(X=0)=1-\int_{\text{support}(y)} Pr(X=0|y)p(y)\;dy=$ $\displaystyle 1-\int_{y=0}^5 \frac{Pr(X=0|y)}{5}\;dy$ $\displaystyle =1-\frac{1}{5}\int_{y=0}^5 e^{-y} \;dy$

CB

can this be more simple?
• Mar 8th 2010, 03:31 AM
CaptainBlack
Quote:

Originally Posted by chialin4
can this be more simple?

The last integral is about as simple as you are going to get.

CB