# Talk:Brownian bridge

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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 (talkcontribs)

Correct. I've fixed it. Michael Hardy 18:52, 5 June 2007 (UTC)
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)
I incorporated a reference where this is proved. --User:zeycus 10:06, 18 June 2007 (UTC)

## 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)

## Lévy bridge?

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

## 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)

You have the distribution for any one point along the bridge, so you can simulate it recursively by first simulating one point in between the first pair, then you have two adjacent pairs and can simulate another two points for those two bridges, etc.. CHF 204.111.91.15 (talk) 18:01, 18 October 2014 (UTC)

## Intuitive Remarks

This is really poorly written. I'm not sure what the second paragraph is trying to get across otherwise I'd try to edit it. Rewrite anyone? Ckhenderson (talk) 22:45, 31 July 2011 (UTC)

## Independence?

In the construction W(t) = B(t) + t Z, it doesn't say that Z must be independent of B. Could someone confirm that independence is required (or at least that certain dependencies are not allowed)? LachlanA (talk) 04:07, 16 November 2011 (UTC)

Yes indeed. This was a mistake. If B is a Brownian bridge, and W(t) = B(t) + t Z is a Brownian motion, then B(t) = W(t) - t W(1) is independent of W(1) = Z. So Z must be independent of B. Sorry to answer so late... --31.39.233.46 (talk) 14:34, 14 August 2015 (UTC)

## Is it a diffusion process?

Is a Brownian bridge a diffusion process, using E.B. Dynkin's definition? If so, where is the proof? Robert O'Rourke (talk) 00:49, 12 June 2012 (UTC)