Theorem: Excess Kurtosis of Poisson Distribution

Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$.
Then the excess kurtosis $\gamma_2$ of $X$ is given by: 
:$\gamma_2 = \dfrac 1 \lambda$

Proof:
From the definition of excess kurtosis, we have: 
:$\gamma_2 = \expect {\paren {\dfrac {X - \mu} \sigma}^4} - 3$
where: 
:$\mu$ is the expectation of $X$. 
:$\sigma$ is the standard deviation of $X$.
By Expectation of Poisson Distribution, we have: 
:$\mu = \lambda$
By Variance of Poisson Distribution, we have: 
:$\sigma = \sqrt {\lambda}$ 
So: 
{{begin-eqn}}
{{eqn | l = \gamma_2
      | r = \dfrac {\expect {X^4} - 4 \mu \expect {X^3} + 6 \mu^2 \expect {X^2} - 3 \mu^4} {\sigma^4} - 3
      | c = Kurtosis in terms of Non-Central Moments
}}
{{eqn | r = \frac {\expect {X^4} - 4 \lambda \paren {\lambda^3 + 3 \lambda^2 + \lambda} + 6 \lambda^2 \paren {\lambda^2 + \lambda} - 3 \lambda^4} {\lambda^2} - 3
      | c = Skewness of Poisson Distribution, Variance of Poisson Distribution: Proof 1
}}
{{end-eqn}}
To calculate $\gamma_2$, we must calculate $\expect {X^4}$.
We find this using the moment generating function of $X$, $M_X$. 
By Moment Generating Function of Poisson Distribution, this is given by: 
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
From Moment in terms of Moment Generating Function:
:$\expect {X^4} = \map {M_X' ' ' '} 0$
In Skewness of Poisson Distribution, it is shown that:
{{begin-eqn}}
{{eqn | l = \map {M_X' ' '} t
      | r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1}^2 e^{\lambda \paren {e^t - 1} + t}
      | c = 
}}
{{eqn | r = \lambda^3 e^{\lambda \paren {e^t - 1} + 3t} + 3 \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda e^{\lambda \paren {e^t - 1} + t}
      | c = expanding
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \map {M_X' ' ' '} t
      | r = \lambda^3 \paren {\lambda e^t + 3} e^{\lambda \paren {e^t - 1} + 3t} + 3 \lambda^2 \paren {\lambda e^t + 2} e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}
      | c = Chain Rule for Derivatives, Derivative of Exponential Function, Derivative of Power
}}
{{eqn | r = \lambda^4 e^{\lambda \paren {e^t - 1} + 4t} + 6 \lambda^3 e^{\lambda \paren {e^t - 1} + 3t} + 7 \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda e^{\lambda \paren {e^t - 1} + t}
      | c = Exponential of Sum
}}
{{end-eqn}}
Setting $t = 0$:
{{begin-eqn}}
{{eqn | l = \expect {X^4}
      | r = \lambda^4 e^{\lambda \paren {e^0 - 1} + 0} + 6 \lambda^3 e^{\lambda \paren {e^0 - 1} + 0} + 7 \lambda^2 e^{\lambda \paren {e^0 - 1} + 0} + \lambda e^{\lambda \paren {e^0 - 1} + 0}
}}
{{eqn | r = \lambda^4 + 6 \lambda^3 + 7 \lambda^2 + \lambda
      | c = Exponential of Zero
}}
{{end-eqn}}
Plugging this result back into our equation above:
{{begin-eqn}}
{{eqn | l = \gamma_2
      | r = \frac {\expect {X^4} - 4 \lambda \paren {\lambda^3 + 3 \lambda^2 + \lambda} + 6 \lambda^2 \paren {\lambda^2 + \lambda} - 3 \lambda^4} {\lambda^2} - 3
      | c = 
}}
{{eqn | r = \frac {\paren {\lambda^4 + 6 \lambda^3 + 7 \lambda^2 + \lambda} - 4 \lambda \paren {\lambda^3 + 3 \lambda^2 + \lambda} + 6 \lambda^2 \paren {\lambda^2 + \lambda} - 3 \lambda^4} {\lambda^2} - 3
      | c = 
}}
{{eqn | r = \frac {\paren {1 - 4 + 6 - 3} \lambda^4 + \paren {6 - 12 + 6} \lambda^3 + \paren {7 - 4} \lambda^2 + \lambda } {\lambda^2} - 3
}}
{{eqn | r = \frac {3 \lambda^2 + \lambda} {\lambda^2} - 3
}}
{{eqn | r = \frac 1 \lambda
}}
{{end-eqn}}
{{qed}}
Category:Poisson Distribution
Category:Kurtosis
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