# y follows a Poisson distribution with rate lambda*x

#### zzzc

##### New Member
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
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
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
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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