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I am an Economics undergraduate who was reading through a textbook on statistical theory.

On one of the questions, I had to find the Variance of $X$ the joint probability distribution,

$f(x,y)=\frac{1}{4^{x+y}}$, where $x$ and $y$ were discrete random variables $x=0,1,2,...$ and $y=0,1,2,...$

When calculating $Var(x)$, and trying to find $E(x^2)$ I got stuck at the summation for $\frac{X^2}{4^X}$ for $0\le X$.

Previously in the part when I calculated $E(x)$, I was able to sum $\frac{X}{4^X}$ using an AGP.

However, when looking at the variance portion, I'm not sure what kind of series this is, and what method I can use to derive an answer.

Any help would be greatly appreciated !!

jimjim
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guavajuice
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1 Answers1

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So, you want the value of

$$S := \sum_{k=0}^\infty \frac{k^2}{4^k}$$

Let us start with the geometric series, with $x \in (-1,1)$:

$$\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$

Take the derivative of this on both sides, multiply by $x$, then do both again. (The derivative of the sum can be taken term-by-term since the series converges absolutely.) You'll get that

$$\sum_{k=0}^\infty k^2 x^k = \frac{x(x+1)}{(1-x)^3}$$

This can be applied to your case if you notice that

$$S = \sum_{k=0}^\infty k^2 \left( \frac 1 4 \right)^k$$

i.e. use $x=1/4$.

PrincessEev
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  • hmm, so I see that you used the basic geometric series and did differentiation to get the answer, may I know how you thought of this? what I can think of is because the original series goes from ( 0 + 1 + (\frac{1}{4}$) + (\frac{4}{16}$) + (\frac{9}{64}$) + ... ) so I noticed that there is a difference in the numerator from +1,+3,+5,+7, thus the common difference in the differences is +2... is that why I need to get the second derivative and manipulate to get a solution? Thanks !! – guavajuice Jul 09 '20 at 09:31
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    It's just something I know from experience. I see $\sum k^2 x^k$, which immediately is similar to the geometric series $\sum x^k$. I think how I might transform the latter into the former; taking the derivative would get me $\sum kx^{k-1}$. That's nice but I have to multiply by $x$ to get closer to the desired form, and I have to do the same again to get the exact same form. – PrincessEev Jul 09 '20 at 18:45