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Lecture 23
1. Functions of a continuous random variable, continued
The problem: Y = g(X); find fY in terms of fX.
The solution: First compute FY , by hand, in terms of FX , and then use the
fact that F ′ Y =^ fY^ and^ F^
′ X =^ fX^.
Example 23.1. Suppose X has density fX. Then let us find the density
function of Y = X^2. Again, we seek to first compute FY. Now, for all a > 0,
FY (a) = P{X 2 ≤ a} = P
a ≤ X ≤
a
= FX
a
− FX
a
Differentiate [d/da] to find that
fY (a) =
fX (
a) + fX (−
a)
2
a
On the other hand, fY (a) = 0 if a ≤ 0. For example, consider the case that
X is standard normal. Then,
fX 2 (a) =
e −a √ 2 πa
if a > 0 ,
0 if a ≤ 0.
Or if X is Cauchy, then
fX 2 (a) =
π
a(1 + a)
if a > 0 ,
0 if a ≤ 0.
Or if X is uniform (0 , 1), then
fX 2 (a) =
a
if 0 < a < 1 ,
0 otherwise.
Example 23.2. Suppose μ ∈ R and σ > 0 are fixed constants, and define
Y = μ + σX. Find the density of Y in terms of that of X. Once again,
F (^) Y (a) = P {μ + σX ≤ a} = P
X ≤
a − μ
σ
= F X
a − μ
σ
Therefore,
fY (a) =
σ
fX
a − μ
σ
For example, if X is standard normal, then
fμ+σX (a) =
2 πσ 2
exp
(x − μ) 2
2 σ 2
This is the socalled N (μ , σ 2 ) density.
Example 23.3. Suppose X is uniformly distributed on (0 , 1), and define
Y =
0 if 0 ≤ X <
1 3 ,
1 if
1 3
≤ X <
2 3
2 if 2 3
≤ X < 1.
Then, Y is a discrete random variable with mass function,
fY (x) =
1 3 if x = 0, 1, or 2,
0 otherwise.
For instance, in order to compute fY (1) we note that
fY (1) = P
≤ X <
1 / 3
fX (y) ︸ ︷︷ ︸ ≡ 1
dy =
Example 23.4. Another common transformation is g(x) = |x|. In this
case, let Y = |X| and note that if a > 0, then
F (^) Y (a) = P{−a < X < a} = F (^) X (a) − F (^) X (−a).
Else, F (^) Y (a) = 0. Therefore,
fY (a) =
fX (a) + fX (−a) if a > 0 ,
0 if a ≤ 0.
