# 7 it should be clear that the independent increments

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Unformatted text preview: these two observations, we see that EN var (Yi ) is the contribution to the variance from the variability of the Yi , while var (N )(EYi )2 is the contribution from the variability of N . Proof. When N = n, S = X1 + · · · + Xn has ES = nEYi . Breaking things down according to the value of N , ES = = 1 X n=0 1 X n=0 E (S |N = n) · P (N = n) nEYi · P (N = n) = EN · EYi For the second formula we note that when N = n, S = X1 + · · · + Xn has var (S ) = n var (Yi ) and hence, E (S 2 |N = n) = n var (Yi ) + (nEYi )2 Computing as before we get ES 2 = = 1 X n=0 1 X n=0 E (S 2 |N = n) · P (N = n) {n · var (Yi ) + n2 (EYi )2 } · P (N = n) = (EN ) · var (Yi ) + EN 2 · (EYi )2 To compute the variance now, we observe that var (S ) = ES 2 (ES )2 = (EN ) · var (Yi ) + EN 2 · (EYi )2 = (EN ) · var (Yi ) + var (N ) · (EYi )2 (EN · EYi )2 where in the last step we have used var (N ) = EN 2 (EN )2 to combine the second and third terms. For part (iii), we note that in the special case of the Poisson, we have EN = and var (N ) = , so the result follows from va...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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