Finding the expected value of discrete random variable with distribution $f(x) =k(2/3)^x$, $x\in\mathbb N$

Question

Let: $$f(x) = \begin{cases} k(\frac23)^x, & \text{if }x \in \{1, 2, 3,\ldots\} \\[2ex] 0 & \text{elsewhere}\end{cases}$$

Find $k$:

Because $x$ is countable, it is discrete. So, a $\sum$ is appropriate, correct? Here is what I did:

$$\sum_{x=1}^\infty \left(\frac23\right)^x=1$$ $$k\sum_{x=1}^\infty \left(\frac23\right)^x=1$$ $$\sum_{x=1}^\infty \left(\frac23\right)^x=2$$

Thus, $$k(2)=1$$ $$k=\frac12$$

Now, find $E(x)$:

$$E(x)=\sum_{x=1}^\infty xf(x)=\frac12 \sum_{x=1}^\infty x \left(\frac23\right)^x$$

$$\sum_{x=1}^\infty x \left(\frac23\right)^x=6$$

So,

$$E(x)=\frac12(6)=3$$

Is this approach correct? Thanks in advance.

Answer 1

^{Summary of comments can work as an answer here.}

Everything is correct. You may need to justify the assertion about $\sum x(2/3)^x$. And it is not expected value of a pdf it is expected value of a random variable. -- André Nicolas

Concerning "how to evaluate the series in a swift manner":

The first is an infinite geometric series (with common ratio $2/3$, so it converges) and its sum is well known. But both are binomial expansions: $(1 - x)^{-1} = \sum x^n = 1 + x + x^2 + \ldots$ and $(1 - x)^{-2} = \sum nx^{n-1} = 1 + 2x + 3x^2 + \ldots$. In the summations you have, $2/3$ is common to all terms, so remove it and you get these forms. -- M. Vinay