What is the Expected Value of X with Poisson Distribution?

  • Thread starter Thread starter ArcanaNoir
  • Start date Start date
  • Tags Tags
    Sum
ArcanaNoir
Messages
778
Reaction score
4

Homework Statement



Finding the expected value of x, with poisson distribution. I don't follow the sum. It goes like this:

E(x)= \sum_{x=0}^{\infty} \frac{xe^{-\lambda}\lambda^x}{x!}
= e^{-\lambda} \sum_{x=0}^{\infty} \frac{x\lambda^x}{x(x-1)!}
= \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!}
= \lambda e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!}
= \lambda e^{-\lambda}e^{\lambda} = \lambda


So basically the part I don't get is why they say
\sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!} = e^{\lambda}
 
Physics news on Phys.org
That is just the Taylor series expansion of e^x. Remember that

f(x)=\sum_{k=0}^{+\infty}{\frac{f^{(k)}(0)}{k!}x^k}

So if f(x)=e^x, then

e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}
 
Thanks micro. This chapter is going to be the death of me. All kinds of crazy sums that end up with specific values that I'm supposed to remember from two weeks in calc II. Doomed! I hate sums.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top