1. DISTRIBUTION THEORY
1.4
CDF transformation
Suppose X is a continuous RV with CDF FX (x), which is increasing over the range of
X. If U = FX (x), then U ∼ U (0, 1).
Proof.
FU (u) = P (U ≤ u)
= P {FX (X) ≤ u}
= P {X ≤ FX−1 (u)}
= FX {FX−1 (u)}
= u,
for 0 < u < 1.
This is simply the CDF of U (0, 1), so the result is proved.
The converse also applies. If U ∼ U (0, 1) and F is any strictly increasing “on its range”
CDF, then X = F −1 (U ) has CDF F (x), i.e.,
FX (x) = P (X ≤ x) = P {F −1 (U ) ≤ x}
= P (U ≤ F (x))
= F (x),
1.5
as required.
Non-monotonic transformations
Theorem 1.3.2 applies to monotonic (either strictly increasing or decreasing) transformations of a continuous RV.
In general, if h(x) is not monotonic, then h(x) may not even be continuous.
For example,
if h(x) = [x],
�
integer part of x
then possible values for Y = h(X) are the integers
⇒ Y is discrete.
16
1. DISTRIBUTION THEORY
However, it can be shown that h(X) is continuous if X is continuous and h(x) is
piecewise monotonic.
1.6
Moments of transformed RVS
Suppose X is a RV and let Y = h(X).
If we want to find E(Y ) , we can proceed as follows:
1. Find the distribution of Y = h(X) using preceeding methods.
�

yp(y)
Y discrete



 y
2. Find E(Y ) =
�


 ∞


yf (y) dy Y continuous
−∞
(that is, forget X ever existed!)
Or use
Theorem. 1.6.1
If X is a RV of either discrete or continuous type and h(x) is any transformation (not
necessarily monotonic), then E{h(X)} (provided it exists) is given by:
�

h(x) p(x)
X discrete



 x
E{h(X)} = �

∞



h(x)f (x) dx X continuous

−∞
Proof.
Not examinable.
Examples:
1. CDF transformation
Suppose U ∼ U (0, 1). How can we transform U to get an Exp(λ) RV?
17
1. DISTRIBUTION THEORY
Solution. Take X = F −1 (U ), where F is the Exp(λ) CDF. Recall F (x) =
1 − e−λx for the Exp(λ) distribution.
To find F −1 (u), we solve for x in F (x) = u,
i.e.,
u = 1 − e−λx
⇒ 1 − u = e−λx
⇒ ln(1 − u) = −λx
− ln(1 − u)
.
λ
− ln(1 − U )
Hence if U ∼ U (0, 1), it follows that X =
∼ Exp(λ).
λ
− ln U
Note: Y =
∼ Exp(λ)
[both (1 − U ) & U have U (0, 1) distribution].
λ
m This type of result is used to generate random numbers. That is, there are good
methods for producing U (0, 1) (pseudo-random) numbers. To obtain Exp(λ) ran− log U
dom numbers, we can just get U (0, 1) numbers and then calculate X =
.
λ
2. Non-monotonic transformations
⇒x=
Suppose Z ∼ N (0, 1) and let X = Z 2 ; h(Z) = Z 2 is not monotonic, so Theorem
1.3.2 does not apply. However we can proceed as follows:
FX (x) = P (X ≤ x)
= P (Z 2 ≤ x)
√
√
= P (− x ≤ Z ≤ x)
√
√
= Φ( x) − Φ(− x),
where
Φ(a) =
�
a
−∞
1
2
√ e−t /2 dt
2π
is the N (0, 1) CDF. That is,
�√ �
d
⇒ fX (x) =
FX (x) = φ x
dx
�
1 −1/2
x
2
18
�
�
�
√
1 −1/2
− φ(− x) − x
,
2