0
$\begingroup$

I have conflicting answers with different methods. Here is the question:

The number of defects per yard in a certain fabric, Y , was known to have a Poisson distribution with parameter $\lambda$. The parameter $\lambda$ was assumed to be a random variable with a density function given by:

$f (\lambda) = e^{−\lambda}, \text{for } \lambda≥0,$

Find the expected number of defects per yard by first finding the conditional expectation of Y for given $\lambda$.

Since $\lambda$ is the expected value of any random variable with a poisson distribution, it follows that $E(Y | \Lambda = \lambda) = \lambda$. Since $Y$ has a poisson distribution, it follows that $E(Y) = E(\lambda) = \lambda$

We have a theorem which states that $E(E(Y | \Lambda = \lambda)) = E(Y).$

$$E(Y) = E(\Lambda) = \int^\infty_0E(Y | \Lambda = \lambda) e^{-\lambda}dλ$$ $$E(Y) = E(\Lambda) = \int^\infty_0\lambda e^{-\lambda}dλ$$ $$E(Y) = E(\Lambda) = 1$$

Thus, $E(Y) = 1$

Find the variance of Y.

It follows that the variance of any random variable with a poisson distribution is $\lambda$.

Thus,$V(Y) = 1$

Using the theorem below, I get the following answer instead $V(Y) = 2$

$V(Y_1)=E[V(Y_1|Y_2)] +V[E(Y_1|Y_2)].$

Applying it to our case, we have

$V(Y)=E[V(Y_1|\Lambda)] +V[E(Y_1|\Lambda)].$

$V(Y)=E[\Lambda] +V[\Lambda].$

$V(Y) = 2$

I am leaning more towards the second answer

$\endgroup$
1
$\begingroup$

1) You stated the theorem, but you didn't use it. It goes like this: $$E[Y] = E\{E[Y|\lambda]\} = E\{\lambda\} = 1.$$

2) \begin{align*} Var[Y] &= Var\{E[Y|\lambda]\}+E\{Var[Y|\lambda]\}\\ &=Var\{\lambda\}+E\{\lambda\}\\ &=1+1\\ &=2 \end{align*} I forgot the name of this property, but this is the one you should use. $Y$ by itself does not necessarily follow a Poisson distribution with rate $\lambda$. However $Y|\lambda$ does, since it was given.

$\endgroup$
  • $\begingroup$ 1. Law of Iterated Expectation, 2. Law of Iterated Variance $\endgroup$ – Graham Kemp Dec 10 '15 at 8:57
  • $\begingroup$ Thanks. I will keep that in mind. $\endgroup$ – Em. Dec 10 '15 at 9:01
1
$\begingroup$

Yes - I too would go with your second answer, for the reason you give.

$Var(Y|\lambda=1)=1$ but $\lambda$ is uncertain so $Var(Y)$ should be bigger than this.

As an empirical test, try this R code:

> cases <- 10000000
> set.seed(2015)
> lambda <- rexp(cases, rate=1)
> y <- rpois(cases, lambda)
> mean(lambda)
[1] 0.9999552
> sd(lambda)^2
[1] 0.9999436
> mean(y)
[1] 1.000106
> sd(y)^2
[1] 2.000172
$\endgroup$
1
$\begingroup$

$Y$ does not have a poison distribution with parameter $1$ so the first answer does not apply.

$Y$ has a conditional poison distribution when given parameter $\lambda$, which itself has an exponential distribution.   So the second method gives the answer. (PS: It's the Law of Iterated Variance.)


For verification, look at the marginal probability mass function of $Y$.

$\begin{align}\mathsf P(Y{=}y) & = \int_0^\infty f_\lambda(h)\; \mathsf P(Y{=}y\mid \lambda{=}h)\operatorname d h\;\mathbf 1_{y\in\Bbb N} \\[1ex] & = \int_0^\infty \frac{e^{-2h}h^y}{y!}\operatorname d h\;\mathbf 1_{y\in\Bbb N} \\[1ex] & = \big(\tfrac 12\big)^{y+1}\;\mathbf 1_{y\in\Bbb N} \end{align}$

$\therefore Y$ has a 0-indexed geometric distribution with success parameter $\tfrac 12$. $$Y\sim\mathcal{Geo}_0(\tfrac 1 2)\\ \mathsf E(Y) = \tfrac {1-\tfrac 12}{\tfrac 1 2}=1\\ \mathsf{Var}(Y)=\tfrac {1-\tfrac 12}{\big(\tfrac 1 2\big)^2}=2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.