By B.V.K.VIJAYAKUMAR, ABHIJIT MAHALANOBIS
ISBN-10: 0511134614
ISBN-13: 9780511134616
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Example text
77). Transformation of random variables If we transform the RV X into a new RV Y according to Y ¼ g(X), then the PDF of Y depends on the PDF of X as well as the mapping g(Á). We can find the PDF of Y by first finding connection between the CDFs and then taking the derivatives to determine the relation between the two PDFs. Let us first consider the case when the mapping g(Á) is one-to-one. , dy ¼ jdg/dxjdx. From this, the relationship between the two PDFs can be seen to be as follows: fX ðxÞ fY ð yÞ ¼ (2:78) dgðxÞ dx À1 x¼g ðyÞ We can use this method for other mappings also.
Mean is 0 and variance is 1) RV Y using Y ¼ aX þ b, with a ¼ 1/X, and b ¼ ÀmX/X. Thus the probability of a Gaussian RV with arbitrary mean and variance taking on values in an interval can be obtained from the error function (which describes probabilities for the unit Gaussian). 4 Expectation Often, we are interested in the average value of an RV X. , mX ¼ Z1 xfX ðxÞdx ¼ E fX g (2:82) À1 where E{X} is known as the expectation of X. Expectation can also be defined for functions of X. Thus, we can obtain the expectation of Y ¼ g(X) as follows.
A Gaussian PDF is one of the few RVs that are completely characterized by their first two moments (namely mean and variance). The Poisson distribution is characterized by a single moment. Moments for uniform PDF For the uniform PDF in Eq. 75), the mean can be easily shown to be (xL þ xR)/2, the mid-point of the interval of support. Its variance can be shown to be ðxR À xL Þ2 =12. 5 from thep random numbers uniformly distributed over [0,1] and multiply the results ffiffiffiffiffi by 12. So far, we have concentrated on a single RV.
Correlation Pattern Recognition by B.V.K.VIJAYAKUMAR, ABHIJIT MAHALANOBIS
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