The Student Room Group

Expectation

In a random walk, suppose I toss an unbiased coin and if it lands head, I get £1, tails £-1.

Let R(i) denote the amount I get in the ith toss.

I have read E(R(i)) = 0, E((R(i))^2) = 0 and E(R(i)R(j)) = 0

I can understand how E(R(i)) = 0 because it's 0.5 * (1) + 0.5 (-1)

What about the other two?

Any help is appreciated.
Reply 1
E(R(i)^2) is most definitely not 0.

For E(R(i)R(j)), if iji \neq j then R(i) and R(j) are independent. So use the theorem that for independent random variables X, Y, E[XY] = E[X]E[Y].
Reply 2
Original post by DFranklin
E(R(i)^2) is most definitely not 0.

For E(R(i)R(j)), if iji \neq j then R(i) and R(j) are independent. So use the theorem that for independent random variables X, Y, E[XY] = E[X]E[Y].


Thanks, I think it's a typo.

E((R(i)^2) = 1
Reply 3
Original post by jamman12345
Thanks, I think it's a typo.

E((R(i)^2) = 1


Yeah that's right. DFranklin has summed up why E(R(i)R(j)) = 0 (due to the memoryless property of a random walk). To see why E(R(i)^2) = 1, you use the identical method you used in working out E(R(i)), except you now want to square the possible values (so the -1 is turned into a +1).
Reply 4
Original post by DFranklin
E(R(i)^2) is most definitely not 0.

For E(R(i)R(j)), if iji \neq j then R(i) and R(j) are independent. So use the theorem that for independent random variables X, Y, E[XY] = E[X]E[Y].


Original post by Noble.
Yeah that's right. DFranklin has summed up why E(R(i)R(j)) = 0 (due to the memoryless property of a random walk). To see why E(R(i)^2) = 1, you use the identical method you used in working out E(R(i)), except you now want to square the possible values (so the -1 is turned into a +1).


Thanks.

My source misquoted, hence my confusion.

The question is actually from an extract of Paul Wilmott's Intro To Quant Finance.

Having flicked through it, he (PW) constantly claims the math required is not that high a level but I suspect he just wants his book to appeal to a wider audience and not put off potential students.

Would you agree, or is the math really not that hard? I mean SDE's are grad level as is measure theory to prove Ito's Lemma non-heuristically, so without a bachelor's math degree, there's no hope in hell of understanding!
Reply 5
Ito's integral may be a large ask, but the original question is really only A-level difficulty.

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