Transforming a Random Variable

Add the same amount to every value

A random variable X takes the values 1, 2 and 3 with probabilities $\frac{1}{4}$1/4, $\frac{1}{2}$1/2 and $\frac{1}{4}$1/4. Its mean is $E\left(X\right)=1\times \frac{1}{4}+2\times \frac{1}{2}+3\times \frac{1}{4}=\frac{1}{4}+1+\frac{3}{4}=2$E(X) = 1 × 1/4 + 2 × 1/2 + 3 × 1/4 = 1/4 + 1 + 3/4 = 2.

Add 2 to every value. The new variable X + 2 takes the values 3, 4 and 5, with the same probabilities $\frac{1}{4}$1/4, $\frac{1}{2}$1/2 and $\frac{1}{4}$1/4. Nothing about the shape changes: the whole distribution slides 2 to the right.

X as four equally likely tickets: one 1, two 2s and one 3. They balance at 2, and the outer two are each 1 away.

X + 2 on the same scale: every ticket has moved 2 to the right, and so has the balance point, now at 4. The outer tickets are still 1 away from it.

The mean moves; the spread does not

The mean slides with the values: $E\left(X+2\right)=3\times \frac{1}{4}+4\times \frac{1}{2}+5\times \frac{1}{4}=\frac{3}{4}+2+\frac{5}{4}=4$E(X + 2) = 3 × 1/4 + 4 × 1/2 + 5 × 1/4 = 3/4 + 2 + 5/4 = 4, which is E(X) + 2. Adding 2 to every value adds 2 to the long-run average.

The distances from the mean do not change. For X they are −1, 0 and 1 from the mean 2. For X + 2 they are −1, 0 and 1 from the mean 4. So the variance is the same: Var$\left(X\right)=1\times \frac{1}{4}+0\times \frac{1}{2}+1\times \frac{1}{4}=\frac{1}{2}$(X) = 1 × 1/4 + 0 × 1/2 + 1 × 1/4 = 1/2, and Var$\left(X+2\right)=\frac{1}{2}$(X + 2) = 1/2 as well.

Multiply every value by 2

Now double every value of X. The variable 2X takes the values 2, 4 and 6, with probabilities $\frac{1}{4}$1/4, $\frac{1}{2}$1/2 and $\frac{1}{4}$1/4. Its mean is $2\times \frac{1}{4}+4\times \frac{1}{2}+6\times \frac{1}{4}=\frac{1}{2}+2+\frac{3}{2}=4$2 × 1/4 + 4 × 1/2 + 6 × 1/4 = 1/2 + 2 + 3/2 = 4, which is 2E(X).

Every distance from the mean doubles too: the distances are −2, 0 and 2 from 4, where before they were −1, 0 and 1 from 2. The variance averages the squared distances, and when a distance doubles, its square is multiplied by $2^2=4$2² = 4: the squared distances are now 4, 0 and 4, where before they were 1, 0 and 1.

So Var$\left(2X\right)=4\times \frac{1}{4}+0\times \frac{1}{2}+4\times \frac{1}{4}=2$(2X) = 4 × 1/4 + 0 × 1/2 + 4 × 1/4 = 2, which is $4\times \frac{1}{2}$4 × 1/2, or 4 × Var(X). The standard deviation is the square root, so it only doubles: from $\sqrt{\frac{1}{2}}\approx 0.71$√(1/2) ≈ 0.71 to $\sqrt{2}\approx 1.41$√2 ≈ 1.41.

2X on the same scale: the tickets are at 2, 4, 4 and 6, balancing at 4. The outer two are now 2 away from the balance point, twice as far as before.

Both at once: aX + b

Multiplying by a and then adding b does both things. The mean follows both steps: E(aX + b) = aE(X) + b. Only the multiplying changes the variance, and it is squared: Var$\left(aX+b\right)=a^2\times$(aX + b) = a² × Var(X). The b does not appear, because a shift does not move any value away from the mean.

For the X above, E(X) = 2 and Var$\left(X\right)=\frac{1}{2}$(X) = 1/2. So E(3X + 1) = 3 × 2 + 1 = 7, and Var$\left(3X+1\right)=3^2\times \frac{1}{2}=\frac{9}{2}=4.5$(3X + 1) = 3² × 1/2 = 9/2 = 4.5.

The standard deviation scales by the size of a, written |a|: $\sigma$σ of aX + b is |a| times $\sigma$σ of X. A negative a reflects the distribution, which does not change its spread: Var$\left(-X\right)={\left(-1\right)}^2\times$(−X) = (−1)² × Var(X) = Var(X).

Changing units is a transformation like this. A temperature in degrees Fahrenheit is F = 1.8C + 32, where C is in degrees Celsius. If C has mean 20 and standard deviation 5, then F has mean 1.8 × 20 + 32 = 68 and standard deviation 1.8 × 5 = 9. Check with the variance: $1.8^2\times 5^2=3.24\times 25=81=9^2$1.8² × 5² = 3.24 × 25 = 81 = 9².

The usual mistakes

Multiplying the variance by a instead of $a^2$a². If Var(X) = 3, then Var(4X) = 16 × 3 = 48, not 4 × 3 = 12. It is the standard deviation that is multiplied by 4.

Adding b to the variance. Var(X + 5) = Var(X): adding 5 to every value moves them all together.

Leaving out a part of the mean. If E(X) = 3, then E(4X + 5) = 4 × 3 + 5 = 17. Dropping the 5 gives 12, and dropping the 4 gives 8.