# Thread: conditioning problem - homework due very soon

1. ## 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?

2. 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?
$Pr(X\ge 1)=1-Pr(X=0)=1-\int_{\text{support}(y)} Pr(X=0|y)p(y)\;dy=$ $1-\int_{y=0}^5 \frac{Pr(X=0|y)}{5}\;dy$ $=1-\frac{1}{5}\int_{y=0}^5 e^{-y} \;dy$

CB

3. But $f_Y(y)=1/5$ on 0<y<5

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

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

$Y\sim U(0,5)$ not $U(0,1)$
Oppss ... working on auto-pilot there.

CB

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

CB
can this be more simple?

6. Originally Posted by chialin4
can this be more simple?
The last integral is about as simple as you are going to get.

CB