<t>It is related, but not quiet. We can think of U(1), U(2)-U(1), and 1-U(2), as the lengths of the three pieces of the stick. Now, max{U(1), U(2)-U(1), 1-U(2)} is nothing but the length of the longest piece, which can range from 1/3 to 1. Note that it is not necessary for the three pieces to form s...

Try solving problems from any good book on probability theory (Sheldon Ross, Chung).

<t>A simpler solution uses conditioning on the first toss. If N represents the number of throws to get a H followed by a T, thenE[N|H]=1+p(1+E[N|H])+(1-p).1, which gives us E[N|H]=2/(1-p)E[N|T]=1+E[N]SoE[N]=pE[N|H]+(1-p)E[N|T] =2p/1-p + (1-p) + (1-p)E[N]E[N].p=[2p+(1-p)^2]/(1-p)E[N]=(p^2+1)/(p(1-p))...

<t>There needs to be at least 1 red ball between each pair of black balls. So the (b-1) spaces between the black balls have to be filled with red balls. That leaves us with a-(b-1) red balls to place. Now they can be placed either in the (b-1) spaces between the black balls, or in the 2 spaces at th...

This ODE can be easily solved by rearranging terms(x+y)dy=(x-y)dyxdy + ydx = xdx - ydyd(xy) = xdx - ydyIntegrating both sides, we getxy = x^2/2 - y^2/2 + c

I realized that the Martingale method does not work here. The key problem is that we don't know that S_z=1, because z is the first time at which the random walk crosses 1. I am sorry for my earlier post, got carried away by the seductive appeal of Martingales!

The idea is inspired by the fact that this is a random walk of a different kind. Martingales rock!!!

This problem can be solved easily by observing that \!\(M_n\ = exp(\S_n-n) is a martingale, and given that z is a stopping time, one can apply Doob's Optional Stopping Theorem to get E[z]=e

X/Y is a Cauchy Distribution only when X & Y are independent normal random variables. Clearly S_n and S_m are not independent. So that counterexample doesn't cut it.