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?
conditional-expectation recursion
asked Nov 30 at 19:29
user603569
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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?
conditional-expectation recursion
asked Nov 30 at 19:29
user603569
728
add a comment |
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?
conditional-expectation recursion
asked Nov 30 at 19:29
user603569
728
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?
conditional-expectation recursion
conditional-expectation recursion
asked Nov 30 at 19:29
user603569
728
asked Nov 30 at 19:29
user603569
728
asked Nov 30 at 19:29
user603569
728
asked Nov 30 at 19:29
user603569
728
asked Nov 30 at 19:29
user603569
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728
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1 Answer
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$E(X_{j+1}|X_j)$ = $X_j * p * (1+f) + X_j * ( 1 - p) * (1-f)$
$E(X_{j+1}|X_j)$ = $X_j *[2pf−f+1]$
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
answered Nov 30 at 19:51
Ofya
5048
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 at 20:04
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
add a comment |
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$E(X_{j+1}|X_j)$ = $X_j * p * (1+f) + X_j * ( 1 - p) * (1-f)$
$E(X_{j+1}|X_j)$ = $X_j *[2pf−f+1]$
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
answered Nov 30 at 19:51
Ofya
5048
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 at 20:04
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
add a comment |
$E(X_{j+1}|X_j)$ = $X_j * p * (1+f) + X_j * ( 1 - p) * (1-f)$
$E(X_{j+1}|X_j)$ = $X_j *[2pf−f+1]$
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
answered Nov 30 at 19:51
Ofya
5048
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 at 20:04
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
add a comment |
$E(X_{j+1}|X_j)$ = $X_j * p * (1+f) + X_j * ( 1 - p) * (1-f)$
$E(X_{j+1}|X_j)$ = $X_j *[2pf−f+1]$
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
answered Nov 30 at 19:51
Ofya
5048
$E(X_{j+1}|X_j)$ = $X_j * p * (1+f) + X_j * ( 1 - p) * (1-f)$
$E(X_{j+1}|X_j)$ = $X_j *[2pf−f+1]$
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
answered Nov 30 at 19:51
Ofya
5048
answered Nov 30 at 19:51
Ofya
5048
answered Nov 30 at 19:51
Ofya
5048
answered Nov 30 at 19:51
Ofya
5048
5048
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 at 20:04
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
add a comment |
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 at 20:04
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
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 at 20:04
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 at 20:04
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
You're welcome, don't forget to like and mark the answer :))
– Ofya
Nov 30 at 20:10
add a comment |
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