conditioning problem - homework due very soon

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March 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?
March 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?

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

CB
March 4th 2010, 10:48 PM
matheagle

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

$Y\sim U(0,5)$ not $U(0,1)$
March 5th 2010, 01:54 AM
CaptainBlack

Quote:

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
March 8th 2010, 03:07 AM
chialin4

Quote:

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?
March 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

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