y follows a Poisson distribution with rate lambda*x

zzzc

New Member
#1
Assume that the number of faults in the roll, say y, follows a Poisson distribution with rate lambda*x, where x is the length of the roll.
Does this mean the pmf is
f(y; lambda*x) = exp(-lambda*x)(lambda*x)^y / y! ?

Thanks
 

zzzc

New Member
#3
Thanks!!

So if lambda follows a Gamma distribution with shape and scale parameters alpha and beta, how do I go about finding the distribution of y (which follows the Poisson distribution with rate lamba*x) ?

Thanks again!
 

BGM

TS Contributor
#4
Given

\( Y|\{\Lambda=\lambda\} \sim \mbox{Poisson}(\lambda x) \)

\( \Lambda \sim \Gamma(\alpha, \beta) \)

Then

\( \Pr\{Y = y\}
= \int_0^{+\infty} \Pr\{Y = y|\Lambda = \lambda\} f_{\Lambda}(\lambda)
d\lambda \)

The conditional p.m.f. and the density is already known.
And you just need to evaluate this integral.
 

zzzc

New Member
#5
So I have
\(y_i \sim Poisson(\lambda x_i)\) and \(\hat{\lambda}_{MLE}={\bar{y}/\bar{x}}\)

Now I need to find \(var(\lambda)\)
Should I have
\(var(\lambda)=var({\bar{y}/\bar{x}})=(E(\bar{y}))^2 var(1/\bar{x})+(E(1/\bar{x}))^2 var(\bar{y})+var(1/\bar{x})var(\bar{y}\)

or should I have
\(\lambda = E(y_i)/x_i \)
\(var(\lambda)=var({y_i/x_i})=(E(y_i))^2 var(1/x_i)+(E(1/x_i))^2 var(y_i)+var(1/x_i)var(y_i) \)

Which one makes sense??
 
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