Translating N (µ, σ 2 ) random variables.
Problem: Suppose X ∼ N (µ, σ 2 ) and that Y = aX + b with a 6= 0. Does Y have a normal distribution? If so, what
is its mean and variance?
Ans. Yes, Y ∼ N (aµ + b, a2 σ 2 ) and here’s why: Y has the correct (cumulative) distribution function F .
A computation will show this: suppose that a > 0 then
F (y) = P (Y ≤ y) = P (aX + b ≤ y) = P
X≤
y−b
a
Z y−b
a
1 x−µ 2
1
e− 2 ( σ ) dx
2πσ −∞
Z y
1 t−(aµ+b) 2
1
=√
e− 2 ( aσ ) dt
2πaσ −∞
=√
t−b
This last equality is obtained via the substitution x =
. The density function f for Y is F 0 which, because of
a
the fundamental theorem of calculus, is very easy to compute:
Z y
1 y−(aµ+b) 2
1 t−(aµ+b) 2
1
1
d
√
e− 2 ( aσ ) dt.
f (y) = F 0 (y) =
e− 2 ( aσ ) dt = √
dy
2πaσ −∞
2πaσ
We recognize f as the p.d.f. of a N (b
µ, σ
b2 ) random variable with µ
b = aµ+b and σ
b2 = (aσ)2 , i.e., Y ∼ N (aµ+b, a2 σ 2 ).
What if a < 0? Let’s see what happens in the above computation for F :
t−b
(for the integral, use the same substitution x =
as above, being careful to remember that a < 0 )
a
F (y) = P (Y ≤ y) = P (aX + b ≤ y)
y−b
=P X≥
(a is negative, so the inequality reverses)
a
Z ∞
1 x−µ 2
1
=√
e− 2 ( σ ) dx
y−b
2πσ a
Z −∞
1 t−(aµ+b) 2
1
=√
e− 2 ( aσ ) dt
2πaσ y
Z y
1 t−(aµ+b) 2
1
√
=
e− 2 ( −aσ ) dt
2π(−aσ) −∞
1 y−(aµ+b) 2
1
e− 2 ( −aσ ) dt and, since −aσ > 0, once again f is the p.d.f. of a
2π(−aσ)
random variable with distribution N (aµ + b, a2 σ 2 ).
As before we have f (y) = F 0 (y) = √
Application: For us, the main use of this result is that it provides an easy way to convert normally distributed
random variables, with arbitray mean µ and variance σ 2 , to a standard normal random variable.
X −µ
More precisely, if you know X ∼ N (µ, σ) and set Z =
then Z ∼ N (aµ + b, a2 σ 2 ) = N (0, 1)
σ
since a = 1/σ and b = −µ/σ.
Table V in your text can now be used to “look up” probabilities involving X, no matter its mean
and variance, by converting them to probabilites for Z.