LECTURE 10
The Bayes variations
Continuous Bayes rule;
Derived distributions
pX|Y (x | y) =
pX,Y (x, y)
pY (y)
�
pY (y) =
• Readings:
Section 3.6; start Section 4.1
x
=
pX (x)pY |X (y | x)
pY (y)
pX (x)pY |X (y | x)
Example:
Review
pX (x)
pX |Y (x | y) =
pX (x) =
�
y
pX,Y (x, y)
pX,Y (x, y)
pY (y)
pX,Y (x, y)
• X = 1, 0: airplane present/not present
• Y = 1, 0: something did/did not register
on radar
fX (x)
fX,Y (x, y)
fX|Y (x | y) =
fX (x) =
� ∞
−∞
fX,Y (x, y)
Continuous counterpart
fY (y)
fX,Y (x, y) dy
fX|Y (x | y) =
fX,Y (x, y)
fY (y) =
FX (x) = P(X ≤ x)
Discrete X, Continuous Y
fY (y ) =
�
x
fY (y)
y
f X ,Y(y,x)=1
1
1
Continuous X, Discrete Y
�
x
fX (x)fY |X (y | x) dx
• It is a PMF or PDF of a function of one
or more random variables with known
probability law. E.g.:
pX (x)fY |X (y | x)
Example:
• X: a discrete signal; “prior” pX (x)
• Y : noisy version of X
• fY |X (y | x): continuous noise model
pY (y) =
fY (y)
What is a derived distribution
pX (x)fY |X (y | x)
fX |Y (x | y) =
x
fX (x)fY |X (y | x)
Example: X: some signal; “prior” fX (x)
Y : noisy version of X
fY |X (y | x): model of the noise
E[X], var(X)
pX |Y (x | y) =
fY (y)
�
=
x
– Obtaining the PDF for
fX (x)pY |X (y | x)
g(X, Y ) = Y /X
pY (y)
involves deriving a distribution.
Note: g(X, Y ) is a random variable
fX (x)pY |X (y | x) dx
Example:
• X: a continuous signal; “prior” fX (x)
(e.g., intensity of light beam);
• Y : discrete r.v. affected by X
(e.g., photon count)
• pY |X (y | x): model of the discrete r.v.
When not to find them
• Don’t need PDF for g(X, Y ) if only want
to compute expected value:
E[g(X, Y )] =
1
� �
g(x, y)fX,Y (x, y) dx dy
How to find them
The continuous case
• Discrete case
• Two-step procedure:
– Obtain probability mass for each
possible value of Y = g(X)
– Get CDF of Y : FY (y) = P(Y ≤ y)
– Differentiate to get
pY (y) = P(g(X) = y)
�
pX (x)
=
fY (y) =
x: g(x)=y
x
dFY
(y)
dy
y
g(x)
.
.
.
.
.
.
.
Example
.
.
.
.
.
.
.
• X: uniform on [0,2]
• Find PDF of Y = X 3
• Solution:
FY (y) = P(Y ≤ y) = P(X 3 ≤ y)
1
= P(X ≤ y 1/3) = y 1/3
2
fY (y) =
Example
dFY
1
(y) =
dy
6y 2/3
The pdf of Y=aX+b
• Joan is driving from Boston to New York.
Her speed is uniformly distributed between 30 and 60 mph. What is the distribution of the duration of the trip?
Y = 2X + 5:
fX
faX
faX+b
200
• Let T (V ) =
.
V
• Find fT (t)
-2
-1
2
3
4
9
f v(v0 )
1/30
fY (y) =
30
60
v0
�
1
y−b
fX
a
|a|
�
• Use this to check that if X is normal,
then Y = aX + b is also normal.
2
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6.041SC Probabilistic Systems Analysis and Applied Probability
Fall 2013
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