Doubt about distribution of the brownian motion 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?
 A: 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.
A: 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

