## Question

Calculate the mean and variance of a Gamma random variable with parameters alpha and lambda.

PDF of Gamma distribution with parameters alpha and lambda is g(x,alpha,lambda)={(1/(lambda^(alpha) Gamma(alpha)) x^(alpha-1)e^(-x/(lambda)) if x>0 quad alpha>0 quad lambda>0),(0 quad otherwise):},

where Gamma(alpha)=int_0^(oo) x^(alpha-1)e^(-x)dx which has following property: Gamma(alpha+1)=alpha Gamma(alpha).

So, mu=E(X)=int_0^(oo) x*1/(lambda^(alpha) Gamma(alpha)) x^(alpha-1)e^(-x/(lambda))=1/(lambda^(alpha) Gamma(alpha)) int_0^(oo)x^(alpha)e^(-x/lambda)dx.

Let y=x/lambda then dy=1/lambda dx or dx=lambda dy. Since x is changing from 0 to oo then y is changing also from 0 to oo.

Thus, integral can be rewritten as 1/(lambda^alpha Gamma(alpha))int_0^(oo) (y lambda)^(alpha) e^(-y) lambda dy=lambda/(Gamma(alpha))int_0^(oo) y^(alpha)e^(-y)dy=(lambda)/(Gamma(alpha)) Gamma(alpha+1)=lambda/(Gamma (alpha))alpha Gamma(alpha)=alpha lambda.

Now, E(X^2)=int_0^(oo) x^2*1/(lambda^(alpha) Gamma(alpha)) x^(alpha-1)e^(-x/(lambda))=1/(lambda^(alpha) Gamma(alpha)) int_0^(oo)x^(alpha+1)e^(-x/lambda)dx.

Let y=x/lambda then dy=1/lambda dx or dx=lambda dy. Since x is changing from 0 to oo then y is changing also from 0 to oo.

Thus, integral can be rewritten as 1/(lambda^alpha Gamma(alpha))int_0^(oo) (y lambda)^(alpha+1) e^(-y) lambda dy=(lambda^2)/(Gamma(alpha))int_0^(oo) y^(alpha+1)e^(-y)dy=

=(lambda^2)/(Gamma(alpha)) Gamma(alpha+2)=(lambda^2)/(Gamma (alpha))(alpha+1) Gamma(alpha+1)=(lambda^2)/(Gamma(alpha))(alpha+1) alpha Gamma(alpha)=alpha(alpha+1) lambda^2.

Finally, Var(X)=E(X^2)-(E(X))^2=alpha(alpha+1)lambda^2-(alpha lambda)^2=alpha lambda^2.

Answer: E(X)=alpha lambda and Var(X)=alpha lambda^2.