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 6.041SC Probabilistic Systems Analysis and Applied Probability Fall 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.