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

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QUESTION:

If XiExp(λ),i=1,2,...,nXiExp(λ),i=1,2,...,n determine the distribution of Zn=X1+...+XnZn=X1+...+Xn 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)=1λexλf(x)=1λexλ          ,x>0,λ>0x>0,λ>0

and have MGF

MX(t)=1(1λt)MX(t)=1(1λt)  or  (1λt)1(1λt)1

X1,...,XnX1,...,Xn is i.i.d then MZn(t)MZn(t) becomes 

MZn(t)=ni=1MXi(t)=MX1(t).MX2(t)...MXn(t)=(MX1(t))n=((1λt)1)n=(1λt)n

MZn(t)=(1λt)n ,t<1λ

The MGF results above are Gamma distribution MGFs with parameters n and λ, or it can be written as ZnGamma(n,λ).

b) Convolution Technique

Zn=X1+...+Xn

Thus

W1=X1+X2W2=W1+X3W3=W2+X4Wn1=Wn2+Xn

By using gz(z)=gx1(x1).gx2(x2)(zx1)dx1 then

For  W1=X1+X2:

xi>0, 0<x1<w1

gw1(w1)=w10gx1(x1).gx2(w1x1)dx1=w101λex1/λ.1λe(w1x1)dx1=w101λ2e(x1+w1x1)/λdx1=w101λ2ew1/λdx1=1λ2ew1/λw101dx1=1λ2ew1/λ(x)|w10=1λ2ew1/λ(w10)=1λ2w1ew1/λuse1=11=1(1)!sothat=1λ2w1ew1/λ.1(1)!=1λ2.(1)!w1ew1/λ

For  W2=W1+X3:

xi>0, 0<w1<w2

gw2(w2)=w20gw1(w1).gx3(w2w1)dw1=w201λ2.(1)!w1ew1/λ.1λe(w2w1)dw1=w201λ3(1)!w1e(w1+w2w1)/λdw1=w201λ3(1)!ew2/λdw1=1λ3(1)!ew2/λw20w1dw1=1λ3(1)!ew2/λ(12w21)|w20=1λ3(1)!ew2/λ(12w220)=1λ31(1)!12w22ew2/λ=1λ3.(2)!w22ew2/λ

based on the iteration above, for Wn1=Wn2+Xn is

gwn1(wn1)=1λn.(n1)!wn1n1ewn1/λ

so

gzn(zn)=1λn.(n1)!zn1nezn/λ

zn>0,n>0,λ>0

The equation is in pdf form with Erlang distribution with parameters λ and n or it can be written as ZnErl(n,λ).

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 (ZnGamma(n,λ)) in the mgf form and the convolution technique produces an Erlang distribution (ZnErl(n,λ)) in the pdf form.