LECTURE 9
Continuous r.v.’s and pdf’s
• Readings: Sections 3.4-3.5
fX(x)
Sample Space
Outline
• PDF review
a
Event {a < X < b }
b
x
• Multiple random variables
– conditioning
– independence
P(a ≤ X ≤ b) =
• Examples
�
x
• E[g(X)] =
fX (x)
FX (x)
xpX (x)
�
E[X]
var(X)
pX,Y (x, y)
a
fX (x) dx
• P(x ≤ X ≤ x + δ) ≈ fX (x) · δ
Summary of concepts
pX (x)
� b
� ∞
−∞
g(x)fX (x) dx
xfX (x) dx
fX,Y (x, y)
pX|A(x)
fX |A(x)
pX |Y (x | y )
fX |Y (x | y)
Joint PDF fX,Y (x, y)
P((X, Y ) ∈ S) =
� �
S
Buffon’s needle
• Parallel lines at distance d
Needle of length � (assume � < d)
• Find P(needle intersects one of the lines)
fX,Y (x, y) dx dy
q
x
l
d
• Interpretation:
P(x ≤ X ≤ x+δ, y ≤ Y ≤ y +δ ) ≈ fX,Y (x, y )·δ 2
• X ∈ [0, d/2]: distance of needle midpoint
to nearest line
• Model: X, Θ uniform, independent
• Expectations:
E[g (X, Y )] =
� ∞ � ∞
−∞ −∞
fX,Θ(x, θ ) =
g (x, y)fX,Y (x, y) dx dy
• Intersect if X ≤
• From the joint to the marginal:
fX (x) · δ ≈ P(x ≤ X ≤ x + δ) =
�
P X≤
• X and Y are called independent if
fX,Y (x, y) = fX (x)fY (y),
0 ≤ x ≤ d/2, 0 ≤ θ ≤ π/2
for all x, y
1
�
sin Θ
2
�
�
sin Θ
2
=
� �
=
4 π/2 (�/2) sin θ
dx dθ
πd 0
0
=
4 π/2 �
2�
sin θ dθ =
2
πd
πd 0
x≤ 2� sin θ
�
�
fX (x)fΘ(θ) dx dθ
�
Conditioning
• Recall
P(x ≤ X ≤ x + δ) ≈ fX (x) · δ
Joint, Marginal and Conditional Densities
• By analogy, would like:
P(x ≤ X ≤ x + δ | Y ≈ y) ≈ fX |Y (x | y) · δ
• This leads us to the definition:
fX |Y (x | y) =
fX,Y (x, y)
Area of slice = Height of marginal
density at x
if fY (y) > 0
fY (y)
• For given y, conditional PDF is a
(normalized) “section” of the joint PDF
Renormalizing slices for
fixed x gives conditional
densities for Y given X = x
Slice through
density surface
for fixed x
• If independent, fX,Y = fX fY , we obtain
Image by MIT OpenCourseWare, adapted from
Probability, by J. Pittman, 1999.
fX |Y (x|y) = fX (x)
Stick-breaking example
• Break a stick of length � twice:
break at X: uniform in [0, 1];
break again at Y , uniform in [0, X]
fX,Y (x, y) =
1
,
�x
0≤y≤x≤�
y
f Y |X (y | x)
f X(x)
L
L
x
y
L
x
fX,Y (x, y) = fX (x)fY |X (y | x) =
on the set:
y
fY (y) =
L
=
=
L
x
E[Y ] =
E[Y | X = x] =
�
yfY |X (y | X = x) dy =
2
� �
0
�
fX,Y (x, y) dx
� �
1
y �x
dx
1
�
log ,
�
y
yfY (y) dy =
0≤y≤�
� �
1
�
�
y log dy =
y
4
0 �
MIT OpenCourseWare
http://ocw.mit.edu
6.041 / 6.431 Probabilistic Systems Analysis and Applied Probability
Fall 2010
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