Recipe

Find the mean after a linear transformation

Use when: you have $Y=aX+b$ and are asked either for $E(Y)$ , or for a missing constant $a$ or $b$ that makes $E(Y)$ hit a target value

Goal: either $E(Y)$ itself, or the specific constant that produces a target mean

Taught in Mean under a linear transformation · a later section extends this recipe with more cases

Quick recipe

Identify $a$ , $b$ , and $E(X)$
Decide which direction you're solving
Apply $E(Y)=aE(X)+b$

1
Write the transformation as $Y=aX+b$ , and read off $a$ (the multiplier) and $b$ (the additive shift) directly from the problem; also note whatever's given about $X$ 's own mean.

Example

$X\sim\operatorname{Normal}(10,2)$ , so $E(X)=10$ ; $Y=3X-4$ gives $a=3$ and $b=-4$ .

Why: every later step is just substituting these three numbers into one formula, so misreading any one of them, especially a sign on $b$ , propagates through the whole problem.
2
Figure out whether $E(Y)$ itself is the unknown (the forward direction), or whether $E(Y)$ is given as a target and one of the constants, $a$ or $b$ , is the actual unknown (the backward direction).
- $E(Y)$ is the unknown — go to Step 3 and substitute directly.
- a target $E(Y)$ is given, and $a$ or $b$ is unknown — go to Step 3 and solve the resulting equation for that constant instead.
- Not yet: the variance of $Y$ under this same transformation — covered in Variance under a linear transformation
Why: both directions use the exact same formula, but misreading which direction a problem asks for leads to solving for the wrong quantity entirely.
3
In the forward direction, substitute the three known numbers straight in and compute. In the backward direction, substitute everything you know, set the expression equal to the target mean, and solve the resulting linear equation for the missing constant.

Example

Forward: continuing Step 1's numbers, $E(Y)=3(10)-4=26$ .

Backward: $X$ has $E(X)=50$ ; find $b$ so that $Y=2X+b$ has $E(Y)=0$ : $0 = 2(50)+b \implies b = -100$

Why: this is the entire content of linearity of expectation, applied to the specific rule $Y=aX+b$ ; nothing about $X$ 's full distribution beyond its mean ever enters.

Common mistakes

Forgetting to add $b$ after multiplying by $a$ , treating the transformation as $Y=aX$ only.
Losing a negative sign on $b$ , especially when $b$ itself is already negative.
Solving the backward direction by plugging the target value into $X$ instead of into $E(Y)$ .
Assuming $E(Y)=aE(X)+b$ needs $X$ to be Normal, or needs any independence condition; linearity of expectation holds unconditionally for any distribution.