Can the expected value of a PMF be zero, as in E[X] = 0?

Question

The whole question is:
Let X be a discrete random variable and let Y = 0.5 X + 3.

(i) Assume that the PMF of X is given by

where k is some suitable constant. Determine the value of k.

(ii) Find E [X] and Var[X].
(iii) Given that X has the PMF found above, find the PMF of Y .
(iv) Without explicit calculation from the PMF of Y , find E [Y ] and V ar[Y ].

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I have to do this for homework but I'm getting E[X] = 0 and can't find variance either.

My progress so far:

Pr(X = -4) + Pr(X = -2) + Pr(X = 0) + Pr(X = 2) + Pr(X = 4) = 1 k * -4^2 + k * -2^2 + k * 0^2 + k * 2^2 + k * 4^2 = 1 16k + 4k + 0 + 4k + 16k = 1 40k = 1 k = 1/40

Distribution table: x | -4 | -2 | 0 | 2 | 4 | Pr(X = x) | 0.4 | 0.1 | 0.0 | 0.1 | 0.4 |
Then: E[X] = -4 * 0.4 + -2 * 0.1 + 0 * 0 + 2 * 0.1 + 4 * 0.4 = -1.6 + -0.2 + 0 + 0.2 + 1.6 = 0

Is this the correct answer till now? If so, how should I proceed with Var[X]?

Any help will be greatly appreciated!

Answer 1

Alternative for finding $\mathbb E(X)$:

$X$ and $-X$ have the same distribution (by definition $X$ is symmetric) so that $\mathbb E(-X)=\mathbb EX$. This can used as follows:

$0=X+(-X)\Rightarrow 0=\mathbb E(X+ (-X))= \mathbb EX+\mathbb E(-X)= 2\times\mathbb EX$ hence $\mathbb EX=0$

In general if $X$ is symmetric and $\mathbb E(X)$ is defined then $\mathbb E(X)=0$.

Answer 2

Since the random variable $X$ takes values $x$ and $-x$ with the same probability (i.e. the pmf of $X$ is symmetric about 0), intuitively, we can expect that its average value will be 0. This can be proved formally using the formula for the expected value:

$E(X) = \sum_x x P(X=x) $

$= \sum_{x < 0} x P(X=x) + \sum_{x>0} xP(X=x)$

$ = \sum_{x >0} (-x) P(X=x) + \sum_{x>0} xP(X=x) = 0$.