Let $\{W_t, t\geq 0\}$ be a standard Brownian motion process. An assignment asks me to calculate the following:
$$\begin{align} &\mathbb{P}(W_4<0),\\
&\mathbb{P}(W_{100} < W_{80}),\\
&\mathbb{P}(W_{100} < W_{80}+2), \\
&\mathbb{P}(W_{3} < W_2 + 2 \text{ and } W_1<0).
\end{align}$$
Now I don't need the answers, I would just like to confirm that my reasoning is correct (since no answers are provided by the course).
For $\mathbb{P}(W_4<0)$, I think that $W_4 \sim N(0,4)?$ For a normal random variable $N(0, \sigma^2)$, we know that the probability distribution function equals
$$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-x^2}{2\sigma^2}}.$$
Then $$\mathbb{P}(W_4 <0) = \int_{-\infty}^0\frac{1}{\sqrt{2\pi\cdot16}}e^{\frac{-x^2}{2\cdot 16}}\quad ?$$
For $\mathbb{P}(W_{100} < W_{80})$, I thought that since $W_{t+s}-W_s \sim N(0,t)$, that
$$\mathbb{P}(W_{100} < W_{80}) = \mathbb{P}(W_{20} <0)$$
which would then be solved in similar fashion to the first probability?
For $\mathbb{P}(W_{100} < W_{80}+2)$, we use the same method as above, only we integrate from $-\infty$ to $2$ instead of $0$?
And for the last, we know that $\mathbb{P}(W_3 < W_2 + 2 \text{ and } W_1<0) = \mathbb{P}(W_3 – W_2 <2 \text{ and } W_1<0) = \mathbb{P}(W_3-W_2 < 2)\mathbb{P}(W_1<0)?$
Is my reasoning correct? Any advise on easier methods or mistakes I've made is very welcome.
Best Answer
Your thoughts are correct… For probabilities of the form $P(X < 0)$ for a centered normal random variable $X$ you can directly conclude $$P(X < 0) = P(X\le 0) = \frac{1}{2}$$ No calculation needed. For the probabilities of the form $P(X < 2)$ you have to do it your way or use tables of a normal distribution