6.041 / 6.431 10. Continuous Bayes Rule & Derived distributions

" LECTURE 10 The Bayes variations Continuous Bayes rule; pX,Y (x, y) pX(x)pY X(y | x) Derived distributions pX|Y (x | y)= pY (y)= pY (|y) Readings: pY (y)= !pX(x)pY X(y | x)• x |Section 3.6; start Section 4.1 Example: Review • X =1, 0: airplane present/not present • Y =1, 0: something did/did not register pX(x) fX(x) on radar pX,Y (x, y) fX,Y (x, y)pX,Y (x, y) fX,Y (x, y) Continuous counterpartpX|Y (x | y)= pY (y) fX|Y (x | y)= fY (y) pX(x)= !pX,Y (x, y) fX(x)= "! fX,Y (x, y) dy fX|Y (x | y)= fX,Y (x, y) fX(x)fY |X(y | x) = y −! fY (y) fY (y) fY (y)= fX(x)fY |X(y | x) dx x FX(x)= P(X # x) Example: X: some signal; “prior” fX(x) E[X], var(X) Y : noisy version of X fY |X(y | x): model of the noise Discrete X, Continuous Y What is a derived distribution pX|Y (x | y)= pX(x)fY |X(y | x) fY (y) • It is a PMF or PDF of a function of one or more random variables with known probability law. E.g.: 1yfY (y)= !pX(x)fY X(y | x) f X,Y(y,x)=1 x |Example: • X: a discrete signal; “prior” pX(x) • Y : noisy version of X • fY |X(y | x): continuous noise model 1x Continuous X, Discrete Y – Obtaining the PDF for fX|Y (x | y)= fX(x)pY |X(y | x) g(X, Y )= Y/X pY (y) " involves deriving a distribution. pY (y)= fX(x)pY |X(y | x) dx Note: g(X, Y ) is a random variable x Example: When not to find them • X: a continuous signal; “prior” fX(x) (e.g., intensity of light beam); • Don’t need PDF for g(X, Y ) if only want • Y : discrete r.v. affected by X to compute expected value: (e.g., photon count) "" pY |X(y | x): model of the discrete r.v. E[g(X, Y )] = 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 – Get CDF of Y : FY (y)= P(Y # y) possible value of Y = g(X)pY (y)= P(g(X)= y) – Differentiate to get= ! pX(x) fY (y)= dFY (y) x: g(x)=y dy Example • X: uniform on [0,2] Find PDF of Y = X3 • Solution:• FY (y)= P(Y # y)= P(X3 # y) = P(X # y 1/3)= 1 y 1/3 2dFY 1 fY (y)= dy (y)=6y2/3 x . ...... y . ...... g(x) Example The pdf of Y=aX+b • Joan is driving from Boston to New York. Her speed is uniformly distributed betwwee 30 and 60 mph. What is the dis-Y =2X + 5: fXtribution of the duration of the trip? faX faX+b 200 Let T (V )= .• V • Find fT (t) -2 -1 23 4 9 v0 fv(v0 ) 30 60 1/30 1 #y − b$ fY (y)= fX |a| a • Use this to check that if X is normal, then Y = aX + b is also normal. MIT OpenCourseWare http://ocw.mit.edu 6.041 /6.431 Probabilistic Systems Analysis and Applied Probability Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.