Question

Calculate the mean and variance of a Poisson random variable with parameter `lambda`.

Answer

Poisson random variable with parameter `lambda` has following probability:

`P(X=x)=(lambda^xe^(-lambda))/(x!)`, `x=0,1,2,...`.

To find its mean and variance we need following fact: `e^(lambda)=sum_(x=0)^(oo)(lambda^x)/(x!)`.

So, `mu=E(X)=sum_(x=0)^(oo)x*(lambda^xe^(-lambda))/(x!)=e^(-lambda)sum_(x=0)^(oo)x*(lambda^x)/(x!)=`

Strip out first summand (when `x=0`):

`=e^(-lambda)(0*(lambda^0)/(0!)+sum_(x=1)^(oo)(x lambda^x)/(x!))=e^(-lambda)sum_(x=1)^(oo)(x lambda^x)/(x!)=e^(-lambda) lambda sum_(x=1)^(oo)(lambda^(x-1))/((x-1)!)=`

Reindex the sum (now `x` starts not from the 1, but from 0):

`=e^(-lambda) lambda sum_(x=0)^(oo) (lambda^x)/(x!)=e^(-lambda) lambda e^( lambda)=lambda`.

Now, `E(X^2)=sum_(x=0)^(oo)x^2*(lambda^xe^(-lambda))/(x!)=e^(-lambda)sum_(x=0)^(oo)x^2*(lambda^x)/(x!)=`

Strip out first summand (when `x=0`):

`=e^(-lambda)(0^2*(lambda^0)/(0!)+sum_(x=1)^(oo)(x^2 lambda^x)/(x!))=e^(-lambda)sum_(x=1)^(oo)(x^2 lambda^x)/(x!)=e^(-lambda) lambda sum_(x=1)^(oo)(x lambda^(x-1))/((x-1)!)=`

Write `x` as `x-1+1`:

`=e^(-lambda) lambda sum_(x=1)^(oo) ((x-1+1) lambda^(x-1))/((x-1)!)=e^(-lambda) lambda sum_(x=1)^(oo) ((x-1) lambda^(x-1))/((x-1)!)+e^(-lambda) lambda sum_(x=1)^(oo) (lambda^(x-1))/((x-1)!)=`

From first sum strip out first summand (when `x=1`), we already found second sum - it equals `lambda`:

`=e^(-lambda) lambda (((1-1)*lambda^(1-1))/((1-1)!)+sum_(x=2)^(oo)((x-1)lambda^(x-1))/((x-1)!))+lambda=e^(-lambda) lambda^2 sum_(x=2)^(oo) (lambda^(x-2))/((x-2)!)+lambda=`

Reindex the sum (now `x` starts not from the 2, but from 0):

`=e^(-lambda) lambda^2 sum_(x=0)^(oo) (lambda^x)/(x!)+lambda=e^(-lambda) lambda^2 e^( lambda)+lambda=lambda^2+lambda`.

Finally, `Var(X)=E(X^2)-(E(X))^2=(lambda^2+lambda)-(lambda)^2=lambda`.

So, both mean an variance equal `lambda`.