I am tasked with the following:
$$E[e^{3B_3}|\mathcal{F}_1]$$
For standard Brownian Motion, i.e. the drift ($\mu$) is $0$, and $\sigma^2=1$.
What I have done so far is the following:
$$E[e^{3B_3}|\mathcal{F}_1]=E[e^{3B_3}|B_1]=E[e^{3B_3-3B_1}\cdot e^{3B_1}|B_1]=e^{3B_1}E[e^{3B_3-3B_1}|B_1]$$
From here, I have tried iterating over possible values of $B_3-B_1$:
$$\rightarrow\sum_{a=0}^\infty e^{3a}P\{B_3-B_1=a|B_1=b\}$$
By independence:
$$\rightarrow\sum_{a=0}^\infty e^{3a}\frac{P\{B_3-B_1=a\}P\{B_1=b\}}{P\{B_1=b\}}=\sum_{a=0}^\infty e^{3a}P\{B_3-B_1=a\}$$
However, I'm kind of stuck here (and, quite frankly, I'm not certain this is the correct method). Any suggestions on this would be sincerely appreciated. Cheers.
You need to compute $\mathbb{E}[e^{3(B_3-B_1)}|B_1]$, but since Brownian motion has independent increments this is the same as just $\mathbb{E}[e^{3(B_3-B_1)}]$. Additionally, you know $B_3-B_1$ is normally distributed with mean $0$ and variance $2$, so using the moment generating function for a normal random variable we compute $$\mathbb{E}[e^{3(B_3-B_1)}] = e^{\frac 12 \cdot 2 \cdot 3} = e^{3}.$$
The approach of trying to sum over all the possible values won't work here because the distributions are continuous, so $P(B_3-B_1 = a) = 0$ for all $a$ because the probability a continuous random variable takes on any particular value is always $0$. You could do it by integrating the pdf for $B_3-B_1$, but you don't have to since the moment generating function for the normal distribution is already known.