# Stochastic Calculus

1. Jan 27, 2005

Hello all

If you throw a head I give you $1. If you throw a tail you give me$1. If $$R_i$$ is the random amount ($1 or -$1) you make on the $$ith$$ toss then why is: $$E[R_i] = 0, E[R^2_i]=1, E[R_iR_j] = 0$$? If $$S_i = \sum^i_{j=1} R_j$$ which represents the total amount of money you have won up to and including the ith toss, then why does $$E[S_i] = 0, E[S_i^2] = E[R_1^2 + 2R_1R_2 + ...] = i$$? I know that if there had been five tosses already then $$E[S_6|R_1,...,R_5] = S_5$$

Any help is appreciated

Thanks

2. Jan 27, 2005

I know that Expected Value is defined as: $$E[X] = \sum_x f(x)P(x)$$ for a discrete variable.

3. Jan 27, 2005

### cepheid

Staff Emeritus
Right, and Ri is what is called (in my probability theory class I'm taking anyways) a Bernoulli Random Variable. There are only two possible outcomes: success, or failure, usually denoted by 1 or 0 respectively. However, since Ri represents the winnings, and they are -1 for tails, the expectation, given your definition is (letting H denote a heads and T a tails):

$$E[X] = \sum_x xp(x) = 1\cdot P(H) -1 \cdot P(T) = 1/2 - 1/2 = 0$$

The expectation can be interpreted as the average value (ie the value expected over time) that the random variable takes on. For a random variable that can only take on the values 1 and -1, what is the average? Since both outcomes have equal weighting (p = 1-p = 1/2), on average you'd expect no net winnings, since the chances of getting a dollar are about equal to that of having to pay up. I hope this gets you started. Now consider the other cases: what is Ri squared? Well, what is 1^2? What is (-1)^2?
ETC...