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Let $B_{t}$ a brownian motion (stochastic process) then I know $B_{t} -B_{s}$ has a normal distribution with mean$=0$ and variance $=t-s$

I want to calculate the following probability: $P(3B_{2}>4)$

I know that if $X \sim N(\mu,\sigma^2$) then if $k$ is an constant $kX \sim N(k \mu,k^2 \sigma^2)$

therefore:

$3B_{2} \sim 3N(0,2)=N(0,2*3^2)=N(0,18)$ then $P(3B_{2}>4)=0.41$

but in $P(B_{2}>4/3)$,
$ B_{2} \sim N(0,2)$ and $P(B_{2}>4/3)=.25$ I don't know what happens, should not give the same probability? on the other hand if $3B_{2} \sim N(0,6)$ the probability of $P(3B_{2}>4)=0.25$ Can andybody helpme?

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Note that if $X\sim N (\mu,\sigma^2)$ then $Pr(X>a)=Pr\left(\frac{X-\mu}{\sigma}>\frac{X-a}{\sigma}\right)$. We devide by the standard deviation to obtain the desired probability using tables for the standard normal distribution.


I think what is going on is that you calculate these probabilities using variance rather than standard deviation. The standard deviation of a Brownian motion is $\sqrt{t}$ and of $3B_t$ is $3\sqrt{t}$. I used MatLab to compute these probabilities and I got: $$Pr(B_2>4/3)\approx 0.1729 \\ Pr(3B_2>4)\approx 0.1729$$ where to compute the first probability I assumed that $B_2\sim N(0,2)$ and to compute the second probability I assumed that $3B_2 \sim N(0,18)$. However, if in computations I pass variances instead to standard deviations then I obtain your mistake.

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  • $\begingroup$ Ohhh I see, thanks! :) $\endgroup$ – Jazz May 15 at 18:59
  • $\begingroup$ You're welcome! $\endgroup$ – Mdoc May 15 at 19:18
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We have $$ \Bbb P(3B_2>4) = \Bbb P\left(\frac {B_2}{\sqrt 2}>\frac{2\sqrt 2}3\right) = \Bbb P\left(\frac {B_2}{\sqrt 2}<-\frac{2\sqrt 2}3\right) = \Phi\left(-\frac{2\sqrt 2}3\right) \ , $$ where $\Phi$ is the repartition of the standard normal distribution $N(0, 1^2)$, since $B_2/\sqrt 2$ respects this distribution. This is roughly, as computed by sage:

sage: T = RealDistribution('gaussian', 1)
sage: T.cum_distribution_function( -2*sqrt(2)/3 )
0.17288929307558015
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