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

discrete mathematicsprobability distributionsrandom variables

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
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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 ].


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!

Best Answer

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 $\mathbb E(X)=0$.

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