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Given a normal random variable X with parameters $\mu$ and $\sigma^2$, find the $E(Y)$ of $Y=aX + b$.

So I started with $E(Y)=E(aX+b)=\frac{1}{\sqrt(2\pi)\sigma}\int_{-\infty}^{\infty}(ax+b)^{{-(ax+b-\mu}^2/2\sigma^2)}$ but this seems a bit unwieldy. Is this the correct approach, and if so, are there any useful substitutions I can make?

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$E[Y] = E[aX+b] = aE[X]+b$ even if $X$ is not a normal random variable; the result holds for all random variables $X$ for which the mean exists. By the way, your integral is wrong: the integrand should be $(ax+b)e^{-(x-\mu)^2/2\sigma^2}$. –  Dilip Sarwate Feb 13 '12 at 2:50
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If you have to do it the "hard way"; it's not even hard. In fact, you don't have to explicitly find any antiderivatives:

Let $c = \frac{1}{\sigma\sqrt{2\pi}}$ .

Then, using the linearity of integration, and noting all the integrals converge:

$$\eqalign{ \Bbb E(aX+b)&=\int_{-\infty}^\infty (ax+b) f_X(x)\,dx \cr &=c\int_{-\infty}^\infty (ax+b) \exp \textstyle({-(x-\mu)^2\over 2\sigma^2})\,dx \cr &= a\cdot c\int_{-\infty}^\infty x \exp {\textstyle({-(x-\mu)^2\over 2\sigma^2}})\,dx +b\cdot c\int_{-\infty}^\infty x \exp (\textstyle{-(x-\mu)^2\over 2\sigma^2})\,dx \cr &=a\,\Bbb E(X)+b\cdot 1\cr &=a\mu+b. } $$

The last integral being $1$, since we are integrating a density function.

It should be remarked this is one way the general formula $\Bbb E(aX+b)=a\Bbb E(X)+b$ can be obtained for continuous $X$. Nothing special about the normal distribution was used.

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It's probably easier to use the general linearity of expectation than to try to integrate. So $$E[Y] = aE[x]+b = a \mu + b$$

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