MAS 209

Distribution of Functions – Change of Variable (Transformation)



Change of Variable (Univariate Case)


Let Y = g(X), where X is a continuous random variable with PDF fX (x), and g
is a strictly monotonic and differentiable function.
Then, the PDF of Y is given by:

d −1
fY (y) = fX g −1 (y) · g (y)
dy

where g −1 is the inverse function of g.
Conditions:
g is one-to-one (strictly monotonic)
g is differentiable, and its inverse g −1 is differentiable




Sam Maseno University August 19, 2025

, Distribution of Functions – Change of Variable (Transformation)



Inverse Transformation Example
Problem. Let X ∼ Exp(1) with PDF
fX (x) = e−x , x > 0.
√
Define Y = X. Find the PDF of Y .

Solution (step-by-step).
Step 1: Identify the transformation and its inverse.
The transformation is
√
Y = g(X) = X, X > 0.
The inverse function is
g −1 (y) = y 2 , y > 0.
Note g is strictly increasing on (0, ∞), so the change-of-variable formula applies.
Step 2: Use the change-of-variable formula.
For y > 0,

d −1 d 2
fY (y) = fX g −1 (y) g (y) = fX (y 2 ) · (y ) .
dy dy
Sam Maseno University August 19, 2025

, Distribution of Functions – Change of Variable (Transformation)




Compute the derivative:
d 2
(y ) = 2y.
dy
Substitute fX (x) = e−x :
2
fY (y) = e−y · 2y, y > 0.

Step 3: Verify the result (normalization).
Check that the PDF integrates to 1:
Z ∞ Z ∞
−y 2
2ye 2
dy let u = y , du = 2y dy = e−u du = 1.
0 0

So the density is valid.

Answer.
2
fY (y) = 2y e−y , y > 0.


Sam Maseno University August 19, 2025