Given a mass distribution for a discrete random variable X, find alpha

Question

The question that I have:

"Let $X$ be a distribution over $\mathbb{N}$ with mass

$P(X = i) = \frac{\alpha}{2^i}$

If $\alpha \in \mathbb{R}$, find $\alpha$ and $E[X]$"

My logic goes as follows:

According to stat axioms, the sum of all probabilities over a sample space $(1,2,3...\infty)$ must be $1$. So:

$1 = \alpha\sum^{\infty}_{i = 1} \frac{1}{2^i}$

Now, this is where I'm not sure I'm correct (pretty sure I'm not):

$\sum^{\infty}_{i = 1} \frac{1}{2^i} = 1$

As such, $1 = \alpha$. After this, I'm not sure how to proceed because:

$E[X] = \alpha \sum_{i=n}^{\infty} \frac{i}{2^i}$

I think the sum converges because $2^i$ grows faster than $i$ but I'm not sure to what number it does.

Answer 1

You have correctly determined that $\alpha = 1$. Then $E X = \sum_{i=1}^\infty i P[X=i] = \sum_{i=1}^\infty i {1 \over 2^i}$.

If $|x|<1$ then $\sum_{k=0}^\infty x^k = {1 \over 1-x}$ and for $x$ inside the radius of convergence we can differentiate to get $\sum_{k=1}^\infty k x^{k-1} = {1 \over (1-x)^2}$, and so $\sum_{k=1}^\infty k x^{k} = {x \over (1-x)^2}$. Setting $x={1 \over 2}$ will give the desired result.

Answer 2

Yes, $\sum_{i=1}^\infty 2^{-i}=1$.   (NB: You are considering $\Bbb N$ to not include $0$.   Mention this in your answer.)

And in general $\sum_{i=1}^\infty r^{-i}= r/(r-1)$ for any $\lvert r\rvert>1$. (Ref: "geometric series".)

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Now since for any positive integer, $i=\sum\limits_{j=1}^i1$, then we have: $$\sum_{i=1}^\infty i2^{-i} =\sum_{i=1}^\infty \left(\sum_{j=1}^i 1\right)2^{-i}$$

The rest is a change of order for summation, and a change of bound variables, and using the above result.