What is the Expected Value of X with Poisson Distribution?

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SUMMARY

The expected value of a random variable X following a Poisson distribution is calculated using the formula E(x) = λ, where λ represents the average rate of occurrence. The derivation involves manipulating the sum E(x) = ∑_{x=0}^{∞} (xe^{−λ}λ^x/x!) to show that it simplifies to λ after applying the Taylor series expansion of e^x. The critical step is recognizing that ∑_{k=0}^{∞} (λ^k/k!) equals e^λ, which is fundamental in deriving the expected value. This discussion highlights the importance of understanding Taylor series in probability theory.

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  • Understanding of Poisson distribution and its properties
  • Familiarity with Taylor series and exponential functions
  • Basic knowledge of calculus, specifically summation techniques
  • Ability to manipulate infinite series and factorials
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ArcanaNoir
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Homework Statement



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

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


So basically the part I don't get is why they say
[tex]\sum_{k=0}^{\infty} \frac{\lambda^{k}}{k!} = e^{\lambda}[/tex]
 
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That is just the Taylor series expansion of [itex]e^x[/itex]. Remember that

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

So if [itex]f(x)=e^x[/itex], then

[tex]e^x=\sum_{k=0}^{+\infty}{\frac{x^k}{k!}}[/tex]
 
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.
 

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