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 Xi∼Exp(λ),i=1,2,...,nXi∼Exp(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)=1λe−xλf(x)=1λe−xλ ,x>0,λ>0x>0,λ>0
and have MGF
MX(t)=1(1−λt)MX(t)=1(1−λt) or (1−λt)−1(1−λt)−1
X1,...,XnX1,...,Xn is i.i.d then MZn(t)MZn(t) becomes
MZn(t)=n∏i=1MXi(t)=MX1(t).MX2(t)...MXn(t)=(MX1(t))n=((1−λt)−1)n=(1−λt)−n
MZn(t)=(1−λt)−n ,t<1λ
The MGF results above are Gamma distribution MGFs with parameters n and λ, or it can be written as Zn∼Gamma(n,λ).
b) Convolution Technique
Zn=X1+...+Xn
Thus
W1=X1+X2W2=W1+X3W3=W2+X4⋮Wn−1=Wn−2+Xn
By using gz(z)=∫∞−∞gx1(x1).gx2(x2)(z−x1)dx1 then
For W1=X1+X2:
xi>0, 0<x1<w1
gw1(w1)=∫w10gx1(x1).gx2(w1−x1)dx1=∫w101λe−x1/λ.1λe−(w1−x1)dx1=∫w101λ2e−(x1+w1−x1)/λdx1=∫w101λ2e−w1/λdx1=1λ2e−w1/λ∫w101dx1=1λ2e−w1/λ(x)|w10=1λ2e−w1/λ(w1−0)=1λ2w1e−w1/λuse1=11=1(1)!sothat=1λ2w1e−w1/λ.1(1)!=1λ2.(1)!w1e−w1/λ
For W2=W1+X3:
xi>0, 0<w1<w2
gw2(w2)=∫w20gw1(w1).gx3(w2−w1)dw1=∫w201λ2.(1)!w1e−w1/λ.1λe−(w2−w1)dw1=∫w201λ3(1)!w1e−(w1+w2−w1)/λdw1=∫w201λ3(1)!e−w2/λdw1=1λ3(1)!e−w2/λ∫w20w1dw1=1λ3(1)!e−w2/λ(12w21)|w20=1λ3(1)!e−w2/λ(12w22−0)=1λ31(1)!12w22e−w2/λ=1λ3.(2)!w22e−w2/λ
⋮
based on the iteration above, for Wn−1=Wn−2+Xn is
gwn−1(wn−1)=1λn.(n−1)!wn−1n−1e−wn−1/λ
so
gzn(zn)=1λn.(n−1)!zn−1ne−zn/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 Zn∼Erl(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 (Zn∼Gamma(n,λ)) in the mgf form and the convolution technique produces an Erlang distribution (Zn∼Erl(n,λ)) in the pdf form.
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