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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 $E(ax)= \frac{1}{a}E(x)$.

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

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

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

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

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    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?2011-11-25
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    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.2011-11-25
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    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.2011-11-25
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    Yea, thank you Srivatsan. x[2011-11-25
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    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.2011-11-26

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