Distribution of a function of a normally distributed variable

Let's say you have a random variable $X$, which is normally distributed according to $X \sim \mathcal{N}(1,2)$. With $1$ being the mean and $2$ being the variance. Now let's say that there is another random variable $Y$, which is defined as $Y = 10X + 5$. How do you now come up with the distribution for $Y$?

For a normal distribution, expected value equals the mean. So for $X$, the expected value is $1$. Can we now say that the $E(Y) = 10 \ E(X) + 5$, and therefore, the expected value of $Y$ is $15$?

Similarly, can we say that the variance for $Y$ is $25$?

Thus, $Y \sim \mathcal{N}(15,25)$?

• As far as I remember, the expected value $E$ is a linear operator, so $E(Y)$ will be $15$ indeed, but the variance is not linear, it is quadratic, so $\mathrm{var}(Y)$ is not $25$. It has something to do with the covariance between $X$ and $Y$, I think, but I'll leave to someone who knows it better to give you a definite answer. Aug 24 '14 at 21:23

The calculation of $E(Y)$ is correct. If $a$ and $b$ are constants, then $E(aX+b)=aE(X)+b$.

For the variance, use the fact that if $a$ andd $b$ are constants, then $$\text{Var}(aX+b)=a^2\text{Var}(X).\tag{1}$$

So the variance of $aX+b$ is independent of $b$. Intuitively, this is because the variance is a measure of variability, so is unchanged if you shift the distribution by a constant.

Remark: Formula (1), and generalizations, will come up quite often in applications.

Note that the expressions for $E(aX+b)$ and $\text{Var}(aX+b)$ are correct whatever distribution $X$ has.

But to get the additional conclusion that $aX+b$ has normal distribution, we need $X$ to have normal distribution.

Note that in the above calculations, we did not show that $Y$ has normal distribution. How to show it depends on the details of how the normal distribution is defined in your course. We will need to use the fact that $\Pr(Y\le y)=\Pr(10X+5\le y)=\Pr\left(X\le \frac{y-5}{10}\right)$.