mean and variance

P(X = k) = 1 / 2^k

k=1,2,...

What will be E(X) and Var(X) ?

 

Here, X has type I geometric distribution with p=q=1/2
So E(X)=1/p=2, Var(X)=q/p-suared=2

 

From first principles?

 

\( E(X)=\sum_{x=1}^{\infty }\frac{x}{2^{x}}
\\now\: see\: in\: numerical\: form 1/2+2/2^2+3/2^3+..............
\\simplified\:as
\\1/2+1/2^2+1/2^3+.............
\\1/2^2+1/2^3+.............
\\1/2^3+...........
\\this\: can\: be \: written\: as
\\\sum_{j=1}^{\infty }\sum_{x=j}^{\infty }\frac{1}{2^{x}}= \sum_{j=1}^{\infty }\frac{1}{2^{j-1}}=2
\\E(X^{2})=\sum_{x=1}^{\infty }\frac{x^{2}}{2^{x}}
\\numerical\: form
\\1/2+2*2/2^2+3*2^3+.............
\\simplified\: as
\\1/2+2/2^2+3/2^3+.......
\\2/2^2+3/2^3+........
\\3/2^3+........
\\so\: now\: you\:know\:value\: 1/2+2/2^2+3/2^3+.......
\\nested\: as\: \sum_{n=1}^{\infty }\left ( \sum_{j=n}^{\infty }\sum_{x=j}^{\infty }\frac{1}{2^{x}}+\frac{n-1}{2^{n-1}} \right )
\\Note:-\frac{n-1}{2^{n-1}} \: is\: added\: because...
I\: give\: you\: a\: example
\\take\: any \: number \: say\: n= 5
\\\sum_{x=5}^{\infty }\frac{x}{2^{x}}=5/2^5+6/2^6+7/2^7+.....
\\=5\left ( 1/2^5+1/2^6+1/2^7+.... \right )+(1/2^6+1/2^7+..)+(...)+...so\: on
\\it\: will\: give\: Sum\: as
\\\sum_{j=5}^{\infty }\sum_{x=j}^{\infty }\frac{1}{2^x}+\frac{4}{2^4}
\\so\: for\: every\: n\: it \: gives\: extra\: \frac{n-1}{2^{n-1}}
\\Hence,
\\E(X^2)=\sum_{n=1}^{\infty }\left ( \sum_{j=n}^{\infty }\sum_{x=j}^{\infty }\frac{1}{2^{x}}+\frac{n-1}{2^{n-1}} \right )=6
\\Var(X)=E(X^2)-(E(X))^2=6-2^2=2
\)
Note:It takes half hour to solve from first principle...... you may choose M.G.F/P.G.F. approach rather.
and yes, quarter hour to write on LaTeX editor

 

Lol, shouldn't they be specifying it is a geometric distribution if we're supposed to use the mean & variance which cannot be worked out from first principles in the exam?
This was an alternative way to solve the question, yet...