# Expected value and Variance of $Y=\frac{1}{a} X-b$ where $X \sim N(\mu, \sigma^2)$

I absolutely know I am not doing this right. :[

Could I get some input or point back in the right direction?

My work done so far is shown below.

Let $X$ be a normal random variable with parameters $N(\mu, \sigma^2)$. Please find the Expected value and Variance of random variable $Y=\frac{1}{a} X-b$, where $a$ and $b$ are constant values.

My work.

\begin{align*} E(Y) &= aE(x)-b = \sum_x \Big(\frac{1}{a} x - b \Big) p_x(x) = \frac{1}{a} \sum_x x p_x(x) - b \\ &= \frac{1}{a} \sum_x x p_x(x) - b \sum_x p_x(x) = a E(x) - b \cdot 1. \end{align*}

If $a = 0$, then $E(x-b) = E(x)$ and if $b = 0$, then $E(ax)= \frac{1}{a}E(x)$.

\begin{align*} \mu &= E(X) = \int_{-\infty}^{\infty} x f(x) dx = \int_{0}^{1} x \Big(\frac{1}{a} X - b \Big) dx = \frac{1}{a} \int_0^1 X^2 - xb ~dx \\ & = \frac{1}{a} \Big( \frac{x^3}{3} - \frac{bx^2}{2} \Big) \text{ from } 1 \text{ to } 0 \\ & = \frac{1}{a} \Big( \frac{1}{3} - \frac{b}{2} \Big) . \end{align*}

$$\text{RV } Y = (X - E(x)^2).$$

$$\sigma^2 = \operatorname{Var}(x) = E[(x) - E[x])^2]$$

$$\operatorname{Var} \Big( \frac{1}{a} X - b \Big) = a^2 \operatorname{Var}(x).$$ $$\int_{\Box}^{\Box}\Big( \frac{1}{a} X - b \Big) - \Big( \frac{1}{a} X - b \Big)^2 \ldots$$

-
Hint: set $1/a = c$. What is $E[cX-b]$? What is the value of $E[cX-b]$ when $X$ is not a normal random variable? – Dilip Sarwate Nov 25 '11 at 22:58
E[cX-b] is similar to E[ax+b] for the discrete RV... I think I have been hooked on the discrete random variable being similar, and am not sure where to go with the actual normal random variable. – Ubez Nov 25 '11 at 23:06
I have typed out the text as in the image. There are several mistakes as it stands. @Ubez Please make appropriate modifications if you are able to find any of the errors. – Srivatsan Nov 25 '11 at 23:11
Yea, thank you Srivatsan. x[ – Ubez Nov 25 '11 at 23:20
I have looked at the original post. Maybe it was typed hurriedly. It is full of errors, some of which ($x$ instead of $X$) point to major difficulties. You seem to be unable to apply the formula $E(aX+b)=aE(X)+b$, which you were given, to the minor variant $E(\frac{1}{a}X-b)$. And there are many other substantial misunderstandings. – André Nicolas Nov 26 '11 at 1:49

Since expectation is linear, you have that $\mathsf{E}(Y)=\mathsf{E}(1/a\:X-b)=1/a\:\mathsf{E}(X)-b$. Furthermore,

\begin{align} \mathsf{Var}(Y) &=\mathsf{E}(Y^{\;2})-\mathsf{E}(Y)^2\\ &=\mathsf{E}(1/a^2X^2-2b/aX+b^2)-(1/a\:\mathsf{E}(X)-b)^2\\ &=1/a^2\mathsf{E}(X^2)-2b/a\mathsf{E}(X)+b^2-1/a^2\mathsf{E}(X)^2+2b/a\mathsf{E}(X)-b^2\\ &=1/a^2\mathsf{E}(X^2)-1/a^2\mathsf{E}(X)^2\\ &=1/a^2\mathsf{Var}(X). \end{align}

-
That is only for discrete random variables though, isn't it? I feel like a normal random variable should be handled differently. Also, this doesn't fit into the E[aX+b] formula because it is a negative b. Sorry, I'm just trying to understand. – Ubez Nov 25 '11 at 23:28
These hold no matter what the distribution of $X$ is. – robjohn Nov 25 '11 at 23:30
Okay, I'll go for it then. Thank you very much. – Ubez Nov 25 '11 at 23:31
@Srivatsan: thanks. I don't know how I missed that. – robjohn Nov 25 '11 at 23:42