0
$\begingroup$

Another stackoverflow question provides the following logic to get from an expected value for Xg(X) where random variable has a Poisson distribution:

\begin{align} E[X g(X)] &= \sum_{x=0}^\infty x g(x) f_X(x) \\ &= \sum_{x=1}^\infty x g(x) f_X(x) & \text{Step 2: addend is zero when $x=0$} \\ &= \sum_{x=1}^\infty x g(x) e^{-\lambda} \frac{\lambda^x}{x!} & \text{Step 3: plug in PMF of Poisson distr.} \\ &= \lambda\sum_{x=1}^\infty g(x) e^{-\lambda} \frac{\lambda^{x-1}}{(x-1)!} & \text{Step 4: rearrange terms} \\ &= \lambda \sum_{x=0}^\infty g(x+1) e^{-\lambda} \frac{\lambda^x}{x!}. & \text{Step 5: shift indexing} \end{align}

What is the logic for getting from Step 3 to Step 4? How can you reindex the f(x) portion, but not the g(x) portion, and where does the lambda constant come from?

(Please note, new user here, so I apparently cannot comment on their original posting for clarification.)

$\endgroup$
  • 2
    $\begingroup$ It's merely that $\frac{x}{x!} = \frac{1}{(x-1)!}.$ One you do that, it is also nice to factor out $\lambda^x$ as $\lambda \cdot \lambda^{x-1},$ since that's yields again the distribution of a Poisson RV (albeit shifted). The reindexing is only occurring in step 5. $\endgroup$ – stochasticboy321 Oct 12 '18 at 19:00
  • $\begingroup$ Thank you! @stochasticboy321 $\endgroup$ – user603569 Oct 12 '18 at 19:06
0
$\begingroup$

Answer posted in comments:

It's merely that x/x!=1/(x−1)! Once you do that, it is also nice to factor out λ^x as λ⋅λ^(x−1) since that's yields again the distribution of a Poisson RV (albeit shifted). The reindexing is only occurring in step 5. – stochasticboy321

(Commenting to mark this as "answered")

$\endgroup$

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.