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I would expect the variance for the general brownian bridge W(T) between T1 and T2 to be (T-T1)(T2-T) / ((T2-T1)^2), rahter than / (T2-T1)

Any thoughts anyone? — Preceding unsigned comment added by 12.42.51.27 (talk • contribs)

Correct. I've fixed it. Michael Hardy 18:52, 5 June 2007 (UTC)[reply]

I see someone reverted back and removed the square exponent, and I completely agree. Otherwise, notice that the variance in any

T\in (T_{1},T_{2})

would be always less that 0.25! --User:zeycus 11:36, 17 June 2007 (UTC)[reply]

I incorporated a reference where this is proved. --User:zeycus 10:06, 18 June 2007 (UTC)[reply]

Variance

The general case variance does not agree with the most common example of a brownian bridge. Bt-tBt in the 0-1 interval. it seems to me that the variance for the general case should be (T-T0)(T1-T)^2/(T2-T1). it is possible to derive the variance of the general process using integration by parts, theorem 4.1.5 in oksendal.

JHS 201.244.172.86 20:48, 5 November 2007 (UTC)[reply]

Lévy bridge?

Is there an analogue of the Brownian bridge, but for a Lévy flight? Albmont (talk) 20:34, 19 November 2007 (UTC)[reply]

simulation of a brownian bridge

It would be nice if there was a comment on how one might simulate a Brownian bridge. Pdbailey (talk) 21:04, 18 February 2008 (UTC)[reply]

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