Standard Brownian Motion: Conditional distribution

I missed a lecture the other day on Brownian motion and am writing a class quiz on Monday. I have been working through some questions but seem to struggle with this one. Apparently it was discussed in the missed lecture.

We have $X$ being standard Brownian motion. Give the distribution of $X(2)$ given that $X(1) = 10$.

So my attempt at it is that I assume the Markov property of Brownian motion. I then get

$$\left[X(2) \in x \mid X(1) = 10 \right] = p(x-10,2-1)$$ From this we assume: $$\left[X(2) \in x \mid X(1) = 10 \right] \sim N(10,1)$$

On the other hand, I wonder whether all X(t) are not independent. I know the increments are independent. Independence would imply that the condition does not affect the distribution though.

Could I please ask whether what I have done seems correct and for you to please guide me in amending what is wrong. Thank you for your time.

• Your calculation is correct. I don't understand your question "I wonder..." though. – Cave Johnson Jul 29 '18 at 13:19
• Thank you very much. I really appreciate your help a lot. I think the "I wonder" is just the pre-quiz self doubt nerves kicking in. I realize my question is probably quite basic and the rules are quite strict here. Would you recommend that I keep or delete the question? – Yes Jul 29 '18 at 13:26
• You are welcome. The question if of good quality and it's absolutely ok if you keep it. – Cave Johnson Jul 29 '18 at 14:08