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I need to find the distribution of $B_s + B_t , \forall \ t,s \geq 0$, where $B$ is a standard Brownian motion.

Here's what I've done:

when $s=t$,

$B_s + B_t = B_t + B_t \sim N(0+0, t+t)=N(0,2t)$

However, the solution combine the $B_t$ and obtain a different variance.

$B_t + B_t = 2B_t \sim N(0,2^2 t)= N(0,4t)$

Shouldn't I be obtaining the same variance regardless of the approach?

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    $\begingroup$ As a hint - is adding two random numbers with the same distribution the same as picking one of them and multiplying it by two? $\endgroup$
    – jbowman
    Dec 7, 2019 at 23:41
  • $\begingroup$ @jbowman so my solution is correct...? $\endgroup$
    – Johnny Ton
    Dec 8, 2019 at 6:26
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    $\begingroup$ Why do you think $B_s = B_t$? They are two different Brownian motions, according to the title of the question. Just because the time stamp is the same doesn't mean their values are. $\endgroup$
    – jbowman
    Dec 8, 2019 at 15:29
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    $\begingroup$ Johnny, suppose someone were to ask about flipping two fair coins and they maintained that because the first coin was heads at flip 17 the other one must also be heads at flip 17 because they are both fair coins. What would you tell them? $\endgroup$
    – whuber ♦
    Dec 8, 2019 at 17:24
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    $\begingroup$ @whuber I think I kinda get it now, thanks! $\endgroup$
    – Johnny Ton
    Dec 9, 2019 at 2:36

1 Answer 1

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Are they two different Brownian motions or the same Brownian motion, just at two different time stamps? Your question title indicates the former. If so, remember that the realization of a B.M. is random. Two different Brownian motions won't have the same sample path. Consequently, if they are different, the variances sum. If you only have one Brownian motion and you are just observing it at time $t$, then the variance is multiplied.by four.

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