If Xi~Exp(λ), i=1,2,...,n, determine the distribution of Zn=X1+...+Xn using a)MGF Technique b)Convolution Technique c)Observe your results in point a)* and b)*, and provide your comments

QUESTION:

If $X_{i}\sim Exp(\lambda ),i=1,2,...,n$ determine the distribution of $Z_{n}=X_{1}+...+X_{n}$ using

a) MGF Technique

b) Convolution Technique

c) Observe your results in point a)* and b)*, and provide your comments

SOLUTION:

a) MGF Technique

pdf of the exponential distribution is

 $f(x)=\frac{1}{\lambda }e^{-\frac{x}{\lambda}}$          ,$x>0,\lambda >0$

and have MGF

$M_{X}(t)=\frac{1}{(1-\lambda t)}$  or  $(1-\lambda t)^{-1}$

$X_{1},...,X_{n}$ is i.i.d then $M_{Z_{n}}(t)$ becomes 

$\begin{aligned}M_{Z_{n}}(t)&=\prod_{i=1}^{n}M_{X_{i}}(t)\\&=M_{X_{1}}(t).M_{X_{2}}(t)...M_{X_{n}}(t)\\&=(M_{X_{1}}(t))^{n}\\&=((1-\lambda t)^{-1})^{n}\\&=(1-\lambda t)^{-n}\end{aligned} $

$\boxed {M_{Z_{n}}(t)=(1-\lambda t)^{-n}}$ $\hspace{0,7cm},t<\frac{1}{\lambda }$

The MGF results above are Gamma distribution MGFs with parameters $n$ and $\lambda$, or it can be written as $Z_{n}\sim Gamma(n,\lambda )$.

b) Convolution Technique

$Z_{n}=X_{1}+...+X_{n}$

Thus

$\begin{aligned}W_{1}&=X_{1}+X_{2}\\W_{2}&=W_{1}+X_{3}\\W_{3}&=W_{2}+X_{4}\\\vdots\\W_{n-1}&=W_{n-2}+X_{n}\end{aligned}$

By using $g_{z}(z)=\int_{-\infty }^{\infty }g_{x_{1}}(x_{1}).g_{x_{2}}(x_{2})(z-x_{1})dx_{1}$ then

For  $W_{1}=X_{1}+X_{2}$:

$x_{i}>0$, $0<x_{1}<w_{1}$

$\begin{aligned}g_{w_{1}}(w_{1})&=\int_{0}^{w_{1}}g_{x_{1}}(x_{1}).g_{x_{2}}(w_{1}-x_{1})dx_{1}\\&=\int_{0}^{w_{1}}\frac{1}{\lambda}e^{-x_{1}/\lambda}.\frac{1}{\lambda}e^{-(w_{1}-x_{1})}dx_{1}\\&=\int_{0}^{w_{1}}\frac{1}{\lambda^2}e^{-(x_{1}+w_{1}-x_{1})/\lambda}dx_{1}\\&=\int_{0}^{w_{1}}\frac{1}{\lambda^2}e^{-w_{1}/\lambda}dx_{1}\\&=\frac{1}{\lambda^2}e^{-w_{1}/\lambda}\int_{0}^{w_{1}}1 \hspace{0,2cm}dx_{1}\\&=\frac{1}{\lambda^2}e^{-w_{1}/\lambda}(x)|_{0}^{w_{1}}\\&=\frac{1}{\lambda^2}e^{-w_{1}/\lambda}(w_{1}-0)\\&=\frac{1}{\lambda^2}w_{1}e^{-w_{1}/\lambda}\\use\hspace{0,2cm}1=\frac{1}{1}&=\frac{1}{(1)!} \hspace{0,2cm}so\hspace{0,2cm}that\\&=\frac{1}{\lambda^2}w_{1}e^{-w_{1}/\lambda}.\frac{1}{(1)!}\\&=\frac{1}{\lambda^2.(1)!}w_{1}e^{-w_{1}/\lambda}\end{aligned}$

For  $W_{2}=W_{1}+X_{3}$:

$x_{i}>0$, $0<w_{1}<w_{2}$

$\begin{aligned}g_{w_{2}}(w_{2})&=\int_{0}^{w_{2}}g_{w_{1}}(w_{1}).g_{x_{3}}(w_{2}-w_{1})dw_{1}\\&=\int_{0}^{w_{2}}\frac{1}{\lambda^2.(1)!}w_{1}e^{-w_{1}/\lambda}.\frac{1}{\lambda}e^{-(w_{2}-w_{1})}dw_{1}\\&=\int_{0}^{w_{2}}\frac{1}{\lambda^3(1)!}w_{1}e^{-(w_{1}+w_{2}-w_{1})/\lambda}dw_{1}\\&=\int_{0}^{w_{2}}\frac{1}{\lambda^3(1)!}e^{-w_{2}/\lambda}dw_{1}\\&=\frac{1}{\lambda^3(1)!}e^{-w_{2}/\lambda}\int_{0}^{w_{2}}w_{1} \hspace{0,2cm}dw_{1}\\&=\frac{1}{\lambda^3(1)!}e^{-w_{2}/\lambda}(\frac{1}{2}w_{1}^2)|_{0}^{w_{2}}\\&=\frac{1}{\lambda^3(1)!}e^{-w_{2}/\lambda}(\frac{1}{2}w_{2}^2-0)\\&=\frac{1}{\lambda^3}\frac{1}{(1)!}\frac{1}{2}w_{2}^2e^{-w_{2}/\lambda}\\&=\frac{1}{\lambda^3.(2)!}w_{2}^2e^{-w_{2}/\lambda}\end{aligned}$

$\vdots $

based on the iteration above, for $W_{n-1}=W_{n-2}+X_{n}$ is

$g_{w_{n-1}}(w_{n-1})=\frac{1}{\lambda^n.(n-1)!}w_{n-1}^{n-1}e^{-w_{n-1}/\lambda}$

so

$\boxed {g_{z_{n}}(z_{n})=\frac{1}{\lambda^n.(n-1)!}z_{n}^{n-1}e^{-z_{n}/\lambda}}$

$z_{n}>0, n>0, \lambda>0$

The equation is in pdf form with Erlang distribution with parameters $\lambda$ and $n$ or it can be written as $Z_{n}\sim Erl(n,\lambda)$.

c) Comments for point a* and b*

The results of the distribution of Zn obtained using the MGF technique and the Convolution technique are different, where the mgf technique produces a Gamma distribution ($Z_{n}\sim Gamma(n,\lambda )$) in the mgf form and the convolution technique produces an Erlang distribution ($Z_{n}\sim Erl(n,\lambda)$) in the pdf form.