4

I'm interested in how to use a Poisson distribution to evaluate $\sum\limits_{x=0}^\infty \frac{(x^2-x+1)(2^x)}{x!}$

I see that this is similar to the general pmf form of $\frac{2^{x}}{x!}$. My strategy is to simplify the summation such that it equals $1$, which is the total probability. To do this, inside the summation I multiply by $e^{-2}$ and outside by $e^2$. In this case the only thing left after the summation is the $e^2$, which is the answer.

When the $x^2-x+1$ is introduced, I don't see a way to remove it from the summation. I'm wondering if there is a different method for solving this, as I can't see any way to make the summation equal $1$.

Tim
  • 85
  • 4
  • why not use ipsative scaling? These are easily implemented and would provide a good approximating, proportionate workaround to the likelihoods you really want. Simply find the maximum value across a summation, divide each unique value by this max, then repercentage to sum up to 100%. Klugey, I know, especially since you seem to want a mathematical solution, but today's kluge is tomorrow's best practice. This would obviate the need for the poisson, but it's of questionable value anyway since probabilities are definitely not integers which the poisson is designed to model – Mike Hunter Jun 13 '15 at 17:50
  • Please add the `self-study` tag and read its [tag wiki](http://stats.stackexchange.com/tags/self-study/info), modifying your question to include what you've tried and where you got stuck. – Glen_b Jun 13 '15 at 18:01

1 Answers1

5

Some guidance and hints ...

Step1: Try a simpler problem first. Can you do $\sum_{x=0}^{\infty} x2^x/x!$ ?

(Hint: simplify the expression, shift the limits of summation / apply a simple change of variable, pull a constant out of the sum, use the approach you mention in your question to bring the right power of $e$ inside the sum and use the fact that the sum of a pmf is 1)

Step2: expand the quadratic into a sum of terms that are easy to do. I suggest doing $\sum (x^2-x)2^x/x!$ following a similar trick to Step 1, and then doing the "1" separately.

Glen_b
  • 257,508
  • 32
  • 553
  • 939
  • 2
    For the first step, I divided the $x$ in the numerator by the denominator, which changed to $(x-1)!$. I then used $y=x-1$, which changed the numerator to $2^{1+y}$. Pulling the $2$ and resolving the sum gave me an answer of $2e^2$. I'll try solving my problem now, I think I see how the method can be applied there. – Tim Jun 13 '15 at 18:50
  • 2
    Tim -- you might see if you can rewrite $x^2-x$ as a product of two terms. – Glen_b Jun 14 '15 at 00:23