Answer to Question #277889 in Statistics and Probability for Johnnie

Question #277889

Let X1,X2,...,Xn be independent and identifically distributed random variables each having pdf



f(x)=q^xP .x=0,1,2....



P+q=1



Let Sn=X1+X2+...+Xn.



Obtain the pgf of Sn and find the mean and variance of Sn

1
Expert's answer
2022-01-11T18:56:01-0500

X1,X2,...,Xn be IID.

f(x)=qxp;,x=0,1,2,...XG(p)E(X)=qp,Var(X)=qp2MX(t)=p1qet,GX(t)=p1qtSn=X1+X2+...+XnGSn(Z)=GX1(Z)GX2(Z)...GXn(Z)=Πi=1nGXi(Z)GX(Z)=E(Zx)=p1qz      [GX(.)pdf of X]f(x)=q^xp;,x=0,1,2,... \\X\sim G(p) \\E(X)=\dfrac qp, Var(X)=\dfrac q{p^2} \\ M_X(t)=\dfrac p{1-qe^t}, G_X(t)=\dfrac p{1-qt} \\S_n=X_1+X_2+...+X_n \\ G_{S_n}(Z)=G_{X_1}(Z)G_{X_2}(Z)...G_{X_n}(Z)=\Pi_{i=1}^nG_{X_i}(Z) \\G_X(Z)=E(Z^x)=\dfrac p{1-qz}\ \ \ \ \ \ [G_X(.)\rightarrow pdf\ of\ X]

So, GSn(Z)=(p1qz)nG_{S_n}(Z)=(\dfrac p{1-qz})^n

E(Sn)=E(X1+X2+...+Xn)=Σi=1nE(Xi)=Σi=1nqp=nqpE(S_n)=E(X_1+X_2+...+X_n) \\=\Sigma_{i=1}^nE(X_i) \\=\Sigma_{i=1}^n \dfrac qp \\=\dfrac {nq}p

Var(Sn)=Var(X1+X2+...+Xn)=Σi=1nVar(Xi)Var(S_n)=Var(X_1+X_2+...+X_n) \\=\Sigma_{i=1}^nVar(X_i) [Since Xi's are IIDs]

=n×qp2=nqp2=n\times \dfrac q{p^2} \\=\dfrac {nq}{p^2}

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