Suppose we have a random variable \(X\) and we would like to know about a random variable \(Y\) that is defined to be \[Y=a+bX\] where \(a\) and \(b\) are constants.
We start out considering the case when \(X\) is discrete and then move to the case when \(X\) is continuous.
Let \(X\) be a discrete random variable with mean \(\mu_x\) and variance \(\sigma_x^2\) and standard deviation \(\sigma_x\) then \(Y=a+bX\) has mean \[\mu_y = a + b \mu_x\] variance \[\sigma_y^2 = b^2 \sigma_x^2\] and standard deviation \[\sigma_y = |b|\sigma_x\]
Let’s check this result with an example.
Consider a random variable \(X\) with the following probability distribution
| \(x\) | \(2\) | \(4\) | \(7\) |
|---|---|---|---|
| \(p(x)\) | \(0.25\) | \(0.35\) | \(0.40\) |
The mean of \(X\) is \[\mu_x=E(X) = 2 \cdot 0.25 + 4 \cdot 0.35 + 7 \cdot 0.40 = 4.7\]
The variance of \(X\) is \[\sigma_x^2 = E(X-\mu_x)^2 = (2-4.7)^2 \cdot 0.25 + (4-4.7)^2 \cdot 0.35 + (7-4.7)^2 \cdot 0.40 = 4.11\]
So the standard deviation of \(X\) is \[\sigma_x = \sqrt{4.11} = 2.027313\]
Now, consider a new random variable \(Y = 3 + 5X\). What is the mean, variance and standard deviation of \(Y\)?
First, let’s find the probability distribution of \(Y\)
| \(y=3+5x\) | \(13\) | \(23\) | \(38\) |
|---|---|---|---|
| \(p(y)=p(x)\) | \(0.25\) | \(0.35\) | \(0.40\) |
And then we can directly compute the mean, variance and standard deviation of \(Y\) and check the formulas presented at the beginning of this section
The mean of \(Y\) is \[\mu_y=E(Y) = 13 \cdot 0.25 + 23 \cdot 0.35 + 38 \cdot 0.40 = 26.5=3+5 \mu_x =3+5\cdot 4.7\] the variance of \(Y\) is \[\sigma_y^2 = E(Y-\mu_y)^2 = (13-26.5)^2 \cdot 0.25 + (23-26.5)^2 \cdot 0.35 + (38-26.5)^2 \cdot 0.40 = 102.75=5^2\sigma_x^2=5^2 \cdot 4.11\] and the standard deviation of \(Y\) is \[\sigma_y = \sqrt{102.75}=10.14=|b|\sigma_x=5 \cdot 2.027313\]
The same result holds for the continuous case. It’s a little harder to show an example for the continuous case since we would have to work with integrals and densities.
In summary, for any random variable \(X\) with mean \(\mu_x\) and standard deviation \(\sigma_x\) and constants \(a\) and \(b\) we have that if \[Y=a+bX\] then the mean of Y is \[\mu_y = E(Y) = a + b \mu_x\] and the variance of Y is \[\sigma_y^2 = b^2 \sigma_x^2\] and the standard deviation of Y is \[\sigma_y = |b| \sigma_x\]
If you are interested, here is an outline of proofs of results for the discrete random variables. For the continuous random variables, we would replace all the sums with integrals.
\(\mu_y = \sum yp(y) = \sum (a+bx)p(x) = \sum ap(x) + \sum bxp(x) = a \sum p(x) + b \sum xp(x) = a + b \mu_x\)
\(\sigma_y^2 = \sum (y-\mu_y)^2 p(y) = \sum (a+bx - (a+b \mu_x))^2p(x)=\sum b^2(x-\mu_x))^2p(x)=b^2\sum (x-\mu_x)^2p(x) = b^2 \sigma_x^2\)