martingale3.pdf - STAT331 Combining Martingales Stochastic...

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STAT331 Combining Martingales, Stochastic Integrals, and Applications to Logrank Test & Cox’s Model Because of Theorem 2.5.1 in Fleming and Harrington, see Unit 11: For count- ing process martingales with continuous compensators, the compensator fully determines the covariance function. Example 12.1: Let N ( t ) = 1( U t, δ = 1) and A ( t ) = t 0 λ ( u ) Y ( u ) du . Suppose that survival time T is continuous and that its hazard function λ ( · ) is bounded. Then A ( · ) is continuous and E ( M 2 ( t )) < for all t. It follows that for s 0, cov( M ( t ) , M ( t + s ))=var( M ( t )) = t 0 λ ( u ) E ( Y ( u )) du = t 0 λ ( u ) P ( U u ) du . In this unit we will discuss functions of martingales. It is easy to show that a linear combination of martingales defined on the same filtration is also a mar- tingale (see Exercises for Unit 11). What about products of martingales? After considering this question, we will show that some stochastic integrals, say Q ( · ) = H ( s ) dM ( s ), with respect to a counting process martingale M are also martingales. These results will then be used to show that the nu- merator of the logrank test and Cox’s score function can be represented as martingales processes. We conclude with a theorem that shows how to find the variance of Q ( t ) as a simple function of the integrand H ( · ) and M ( · ). We begin with a theorem that will enable us to evaluate the covariance be- tween 2 martingale processes. Theorem 12.1: Suppose that M 1 ( · ) and M 2 ( · ) are martingales defined on the same filtration, and that for every t , E ( M j ( t )) = 0 and E ( M 2 j ( t )) < , for j=1,2. Then there exists a right-continuous predictable process < M 1 , M 2 > ( · ) such that M ( · ) def = M 1 ( · ) M 2 ( · ) - < M 1 , M 2 > ( · ) is a zero-mean martingale. 1
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Proof: (adapted from F&H, Thm. 1.4.2). Note that M 1 ( · ) + M 2 ( · ) and M 1 ( · ) - M 2 ( · ) are zero-mean martingales, so that ( M 1 ( · )+ M 2 ( · )) 2 and ( M 1 ( · ) - M 2 ( · )) 2 are sub-martingales (via Jensen’s inequality, see Unit 11 page 11, or directly). Thus, by the Doob-Meyer decomposition, there exist predictable, right-continuous < M 1 + M 2 , M 1 + M 2 > ( · ) and < M 1 - M 2 , M 1 - M 2 > ( · ) such that M ( M 1 + M 2 ) ( · ) def = ( M 1 ( · ) + M 2 ( · )) 2 - < M 1 + M 2 , M 1 + M 2 > ( · ) and (12 . 1) M ( M 1 - M 2 ) ( · ) def = ( M 1 ( · ) - M 2 ( · )) 2 - < M 1 - M 2 , M 1 - M 2 > ( · ) (12 . 2) are martingales. We can choose these martingales to have zero mean. Now define < M 1 , M 2 > ( · ) = <M 1 + M 2 ,M 1 + M 2 > ( · ) - <M 1 - M 2 ,M 1 - M 2 > ( · ) 4 and M ( · ) = M 1 ( · ) M 2 ( · ) - < M 1 , M 2 > ( · ). Subtracting (12.2) from (12.1) and dividing by 4 yields 1 4 ( M ( M 1 + M 2 ) ( · ) - M ( M 1 - M 2 ) ( · ) ) = . . . = M 1 ( · ) M 2 ( · ) - < M 1 , M 2 > ( · ) . Since both of the terms on the left-hand side are zero-mean martingales de- fined on the same filtration, so is M ( · ). 2
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Note: Note that this is not simply an application of the Doob-Meyer decom- position, since M 1 ( · ) M 2 ( · ) is not in general a submartingale. Note (compare with Exercise 6 Unit 10): cov( M 1 ( t ) , M 2 ( t + s )) = E ( < M 1 , M 2 > ( t )) . Thus, knowing < M 1 , M 2 > tells us about cov( M 1 ( · ) , M 2 ( · )).
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