0

I am trying to find the kurtosis of the Poisson - I found

E(X 4)) = λ4 + 6λ3 + 7λ2 + λ`

The 4th central moment I calculated with the binomial expansion:

E(X 4)) - 4 E(X 3)) E(X) + 6E(X 2)) E(X)2 - 4E(X)E(X)3 + E(X)4

= λ4 + 6λ3 + 7λ2 + λ - 4λ(λ3 + 3λ2 + λ) + 6λ2(λ2 + λ) - 4λ4 + λ4

However, I got 4λ2/λ2 when dividing it with sigma4 - according to a solution I got I should be 1/λ.

Heike
  • 49

1 Answers1

1

Your calculation of $\mathsf E[X^3]$ is not correct. Another answer derives it as

$$\mathsf E[X^3]=\lambda^3 + \color{red}3\lambda^2 + \lambda$$

msm
  • 7,147
  • Didn't copy it correctly - I calculated it with your calculation and got the abovementioned result. – Heike Sep 22 '18 at 08:56
  • 1
    Your last calculations is also wrong. It is $3\lambda^2 + \lambda$. Then when you divide it by $\lambda^2$ it becomes $3+\lambda^{-1}$, and if you subtract $3$ it gives you the excess Kurtosis, which is $\lambda^{-1}$. – msm Sep 22 '18 at 11:09