lecture08

# lecture08 - 16.322 Stochastic Estimation and Control, Fall...

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16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Lecture 8 Last time: Multi-dimensional normal distribution 1 x e x p 2 ( x x ) M ( x x ) f () = 1 T 1 n ( 2 π ) 2 M If a set of random variables X i having the multidimensional normal distribution is uncorrelated (the covariance matrix is diagonal), they are independent. The 2 argument of the exponential becomes the sum over i of x i . Thus, the 2 distribution becomes a product of exponential terms in i . If XY = 0 23 2 X Y = X Y 3 = 0 r 2 r n r 1 The general moment of a multidimensional normal distribution, E XX 2 ... X n , 1 is known. Laning and Battin. Random Processes in Automatic Control. The Exponential Distribution Many components, especially electronic components, display constant percentage failure rates over long intervals. Relative rate of failure vs time The familiar “bathtub” curve. Failure rate = E (relative rate of failure) Using the random variable T for time to failure, and assuming a constant failure rate λ (independent of time t ) implies ( T | t ) P t <≤ t + dt T >= dt Page 1 of 6

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16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde This relation alone defines a distribution for time to failure. ( T | t ) ( Tt + d t AND Tt ) P <≤ > Pt t + d t T >= ( PT > t ) ft d t λ dt = () t 1 f ( ττ ) d 0 t ft ) =− f ( ( ( λλ τ ) d τ Integral equation for ) 0 t df f ( t ) Differential equation for f ( t ) dt = c e t t c d c e = 0 0 = 1 c = t = e To find the moments of the distribution, start with the characteristic function.
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## This note was uploaded on 11/07/2011 for the course AERO 16.322 taught by Professor Wallacevandervelde during the Fall '04 term at MIT.

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lecture08 - 16.322 Stochastic Estimation and Control, Fall...

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