Problem where I am getting a negative variance

Question

In Finan's Probability Book 17.20:

Let $X$ be a random variable. Define $Z= {X−E(X)\over σ_X}$. Find $E(Z)$ and $\operatorname{Var}(Z)$.

If I am understanding correctly, ${X−E(X)\over σ_X}={X−E(X)\over\sqrt{E(X^2)-[E(X)]^2}}$

So the variance should be $E\left({X−E(X)\over\sqrt{E(X^2)-[E(X)]^2}}\right)^2-\left[E\left({X−E(X)\over\sqrt{E(X^2)-{[E(X)]}^2}}\right)\right]^2$

which should equal to

$$\left({E(X)^2−2E(X)^2+[E(X)]^2\over{E(X^2)-{[E(X)]}^2}}\right)-\left({[E(X)−E(X)]^2\over{E(X^2)-{[E(X)]}^2}}\right) = -1-0$$

but the answer for the variance is $1$, and I have heard that a negative variance is impossible anyways, so which step did I get wrong?

Answer 1

You made a mistake when you expanded the brackets.

To make it simpler: $$\operatorname{Var}(\frac{X-E(X)}{\sigma(X)})= \frac{\operatorname{Var}(X-E(X))}{\operatorname{Var}(X)} = \frac{E(X^2-2XE(X)+E(X)^2)-E(X-E(X))^2}{E(X^2)-E(X)^2}. $$

Here, $E(X-E(X))=0,$ so $E(X-E(X))^2=0.$

Furthermore, $E(X^2-2XE(X)+E(X)^2)= E(X^2)-2E(X)^2+E(X)^2= E(X^2)-E(X)^2.$

So the desired variance is $$\frac{E(X^2)-E(X)^2}{E(X^2)-E(X)^2}=1.$$

Answer 2

$Z = \frac {X-E[X]}{\sigma_X}= \frac {X}{\sigma_X}-\frac {E[X]}{\sigma_X}\\ E[Z]= E[\frac {X}{\sigma_X}-\frac{E[X]}{\sigma_X}] = \frac {E[X]}{\sigma_X} - \frac {E[X]}{\sigma_X} = 0$

Which means that $var(Z) = E[Z^2]-E[Z]^2 = E[Z^2] = E[(\frac {X}{\sigma_X} - \frac {E[X]}{\sigma_X})^2]= \frac {1}{\sigma^2_X}E[(X - E[X])^2] = \frac {\sigma^2_X}{\sigma^2_X} = 1$

As for your work... $E\left({X−E(X)\over\sqrt{E(X^2)-{[E(X)]}^2}}\right)^2=E\left({X^2−2XE(X)+E(X)^2\over E(X^2)+{[E(X)]}^2}\right)\ne\left({E(X)^2−2E(X)^2+[E(X)]^2\over{E(X^2)-{[E(X)]}^2}}\right)$