Understanding Stochastic Calculus and Expected Value Formulas

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The discussion centers on the expected values associated with a series of coin tosses, where heads yield $1 and tails result in a loss of $1. The expected value of the random variable R_i, representing the outcome of each toss, is calculated to be zero due to equal probabilities of heads and tails. The total winnings S_i after i tosses also has an expected value of zero, while the expected value of S_i squared equals i, reflecting the cumulative variance of the outcomes. The random variable R_i is identified as a Bernoulli Random Variable, with outcomes of 1 and -1, leading to the conclusion that, over time, the average winnings will balance out to zero. This understanding of expected value and variance is crucial in stochastic calculus.
courtrigrad
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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
 
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I know that Expected Value is defined as: E[X] = \sum_x f(x)P(x) for a discrete variable.
 
courtrigrad said:
I know that Expected Value is defined as: E[X] = \sum_x f(x)P(x) for a discrete variable.

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...
 
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