MIT OpenCourseWare http://ocw.mit.edu 18.443 Statistics for Applications Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.−2�2 −�2 + ν2 −2(�2 + ν2) just the function of t we wanted. The constant multiplier, takinat were left aside, is 1 AB = �2��ν 2��ν ��2 + ν2 1 = �, 2�(�2 + ν2) also the correct normalizing constant (as it would have to be, sinc� � � � Fact. If X and Y are independent random variables, X is N (µ, �2) and Y is N(�, ν2), then X + Y is N(µ + �, �2 + ν2). Note. For any two random variables X and Y with finite means (independent or not), E(X + Y ) = EX + EY . And, for any two random variables X and Y with E(X2) < �, E(Y 2) < �, and Cov(X, Y ) = 0, for example, if X and Y are independent, we have Var(X + Y ) = Var(X)+Var(Y ). So, if X + Y has a normal distribution, it must have the given mean and variance. Proof. Clearly, X − µ has a N(0, �2) distribution and likewise Y − � has a N(0, ν 2) distribution. If we can show that X + Y − µ − � has a N(0, �2 + ν2) distribution, it will follow that X + Y has a N(µ + �, �2 + ν2) distribution. So we can assume that µ = � = 0. uRecall that exp(u) means e . The convolution of the N (0, �2) and N(0, ν2) densities, omitting the constant factor A = 1/(2��ν), is � +� � 2 � (t − y)2 yh(t) = exp − 2�2 − 2ν2 dy. −� We can bring a factor exp(−t2/(2�2)) not depending on y outside the integral. The re-maining expression inside the integral, whose exponential is taken, if put over a common denominator, becomes −((�2 + ν2)y2 − 2tν2y)/(2�2ν2). Completing the square, then sub-tracting a term to compensate, this becomes −(�2 + ν2)[(y − v)2 − v2] 2�2ν2 where v = ν2t/(�2 + ν2). Then (�2 + ν2)v2 ν2t2 exp = exp2�2ν2 2�2(�2 + ν2) and we can bring this factor outside the integral because it doesn’t depend on y. Then, the value of the remaining integral doesn’t depend on v and so doesn’t depend on t; it’s a constant B depending on � and ν, specifically, B = �2��ν/��2 + ν2 . The function of t we wind up with, leaving aside such constant multiples, is � � �� � � t2 ν2 t2 exp 1 = exp . This is g the product of those thwhich is e the convolution of two probability densities is a probability density). The proof is complete.
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