7 it should be clear that the independent increments

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online