# Expected value in recursive random variables

Working through some probability problems from Introduction to Probability, Blitzstein:

Kelly makes a series of n bets, each of which she has probability p of winning, independently. Initially, she has x0 dollars. Let Xj be the amount she has immediately after her jth bet is settled. Let f be a constant in (0, 1), called the betting fraction. On each bet, Kelly wagers a fraction f of her wealth, and then she either wins or loses that amount. For example, if her current wealth is 100 dollars and f = 0.25, then she bets 25 dollars and either gains or loses that amount. (A famous choice when p > 1/2 is f = 2p − 1, which is known as the Kelly criterion.) Find E($$X_n$$) (in terms of n, p, f, x0).

Hint: First find $$E(X_{j+1}|X_j)$$.

How can I use the hint? I found $$E(X_1|X_0) = x_0(2pf-f+1)$$, by plugging in the probability of winning (p) * winnings ($$x_0+fx_0$$) plus the probability of losing (1-p) * new winnings ($$x_0-fx_0$$)

But I do not know how to turn this into a recursive formula; any hints on the hint?

$$E(X_{j+1}|X_j)$$ = $$X_j * p * (1+f) + X_j * ( 1 - p) * (1-f)$$

$$E(X_{j+1}|X_j)$$ = $$X_j *[2pf−f+1]$$

now given that we can recursively write

$$E(X_{j+1}|X_j, X_{j-1})$$ = $$X_j *[2pf−f+1]$$

= $$X_{j-1} *[2pf−f+1] *[2pf−f+1]$$

= $$X_{j-1} *[2pf−f+1]^2$$

If we keep doing it we will get:

$$E(X_n)$$ = $$X_0 * [2pf−f+1] ^ n$$

The Kelly Criterion mentioned in the question maximizes the logarithm of this expected value.

Hope it helps

• Yes it does - I got stuck on subscripts for my random variables, but also have a hard time conceptualizing recursive formulas. This makes sense though. Thanks @ofya – user603569 Nov 30 '18 at 20:04
• You're welcome, don't forget to like and mark the answer :)) – Ofya Nov 30 '18 at 20:10