compound interest when interest is a random variable
Say you are investing, and you on average get $2$% interest per bet, with a standard deviation of $3$%.
How can I get, within a confidence interval, an average amount I will have after 100 bets, if interest compounds?
probability
asked 12 hours ago
Frank
15510
add a comment |
Say you are investing, and you on average get $2$% interest per bet, with a standard deviation of $3$%.
How can I get, within a confidence interval, an average amount I will have after 100 bets, if interest compounds?
probability
asked 12 hours ago
Frank
15510
Interesting question. Maybe you can take the logarithm of the product, which will turn it into a sum. Then you have a sum of random variables ... Anyway, just a thought.
– Matti P.
12 hours ago
I am considering running 10,000 simulations in matlab for 10 year predictions and just taking probabilities. I might be able to work out a confidence interval like that if I just find the expectation of 10,000 simulations. Hoping for a real analytical solution though.
– Frank
12 hours ago
add a comment |
Say you are investing, and you on average get $2$% interest per bet, with a standard deviation of $3$%.
How can I get, within a confidence interval, an average amount I will have after 100 bets, if interest compounds?
probability
asked 12 hours ago
Frank
15510
Say you are investing, and you on average get $2$% interest per bet, with a standard deviation of $3$%.
How can I get, within a confidence interval, an average amount I will have after 100 bets, if interest compounds?
probability
probability
asked 12 hours ago
Frank
15510
asked 12 hours ago
Frank
15510
asked 12 hours ago
Frank
15510
asked 12 hours ago
Frank
15510
asked 12 hours ago
Frank
15510
15510
Interesting question. Maybe you can take the logarithm of the product, which will turn it into a sum. Then you have a sum of random variables ... Anyway, just a thought.
– Matti P.
12 hours ago
I am considering running 10,000 simulations in matlab for 10 year predictions and just taking probabilities. I might be able to work out a confidence interval like that if I just find the expectation of 10,000 simulations. Hoping for a real analytical solution though.
– Frank
12 hours ago
add a comment |
Interesting question. Maybe you can take the logarithm of the product, which will turn it into a sum. Then you have a sum of random variables ... Anyway, just a thought.
– Matti P.
12 hours ago
I am considering running 10,000 simulations in matlab for 10 year predictions and just taking probabilities. I might be able to work out a confidence interval like that if I just find the expectation of 10,000 simulations. Hoping for a real analytical solution though.
– Frank
12 hours ago
Interesting question. Maybe you can take the logarithm of the product, which will turn it into a sum. Then you have a sum of random variables ... Anyway, just a thought.
– Matti P.
12 hours ago
Interesting question. Maybe you can take the logarithm of the product, which will turn it into a sum. Then you have a sum of random variables ... Anyway, just a thought.
– Matti P.
12 hours ago
I am considering running 10,000 simulations in matlab for 10 year predictions and just taking probabilities. I might be able to work out a confidence interval like that if I just find the expectation of 10,000 simulations. Hoping for a real analytical solution though.
– Frank
12 hours ago
I am considering running 10,000 simulations in matlab for 10 year predictions and just taking probabilities. I might be able to work out a confidence interval like that if I just find the expectation of 10,000 simulations. Hoping for a real analytical solution though.
– Frank
12 hours ago
add a comment |
1 Answer
1
active
oldest
votes
This is actually a stochastic differential equation. See https://en.wikipedia.org/wiki/Geometric_Brownian_motion.
Say your bet starts at $S_0$, and we model it as $S_t$ over time. Then we can write
$$dS_{t}=mu S_{t},dt+sigma S_{t},dW_{t}$$
where $mu=.02$, $sigma=.03$. Here $W$ represents random brownian motion, so $dW$ is simply a draw from the standard normal. So letting each $dt = 1$, your interest rate is $dS_{t}=mu+sigma dW_{t}$. The analytic solution to this equation is
$$S_{t}=S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma W_{t}right)$$
The pdf of Brownian Motion $W_t$ is Normal with mean $0$ and variance $t$
$${displaystyle f_{W_{t}}(x)={frac {1}{sqrt {2pi t}}}e^{-x^{2}/(2t)}}$$
We want the pdf of $S_t$, $f_{S_{t}}(x)$. Because $S_{t} = g(W_{t})$ with
$$g(x) = S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma xright)$$
we can use the change of variable formula
$$f_{S_{t}}(x)=f_{W_{t}}(g^{-1}(x))|frac{dg^{-1}(x)}{dx}|$$
We have
$$g^{-1}(x) = frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma}$$
and
$$frac{dg^{-1}(x)}{dx} = frac{1}{sigma y}$$
Thus
$$f_{S_{t}}(x) = frac{1}{sigma y sqrt{2pi t}} exp{frac{-(frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma})^2}{2t}}$$
This is the general solution, plug in your values of $t=100$, $mu=.02$, and $sigma=.03$ and you have the pdf for your question!
Funnily enough after all that I'm not sure how to analytically find the mean and standard deviation of $f_{S_{t}}(x)$. But numerically it has mean $approx 7.244S_0$ and standard deviation $approx 2.176S_0$.
answered 11 hours ago
Erik Parkinson
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is actually a stochastic differential equation. See https://en.wikipedia.org/wiki/Geometric_Brownian_motion.
Say your bet starts at $S_0$, and we model it as $S_t$ over time. Then we can write
$$dS_{t}=mu S_{t},dt+sigma S_{t},dW_{t}$$
where $mu=.02$, $sigma=.03$. Here $W$ represents random brownian motion, so $dW$ is simply a draw from the standard normal. So letting each $dt = 1$, your interest rate is $dS_{t}=mu+sigma dW_{t}$. The analytic solution to this equation is
$$S_{t}=S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma W_{t}right)$$
The pdf of Brownian Motion $W_t$ is Normal with mean $0$ and variance $t$
$${displaystyle f_{W_{t}}(x)={frac {1}{sqrt {2pi t}}}e^{-x^{2}/(2t)}}$$
We want the pdf of $S_t$, $f_{S_{t}}(x)$. Because $S_{t} = g(W_{t})$ with
$$g(x) = S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma xright)$$
we can use the change of variable formula
$$f_{S_{t}}(x)=f_{W_{t}}(g^{-1}(x))|frac{dg^{-1}(x)}{dx}|$$
We have
$$g^{-1}(x) = frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma}$$
and
$$frac{dg^{-1}(x)}{dx} = frac{1}{sigma y}$$
Thus
$$f_{S_{t}}(x) = frac{1}{sigma y sqrt{2pi t}} exp{frac{-(frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma})^2}{2t}}$$
This is the general solution, plug in your values of $t=100$, $mu=.02$, and $sigma=.03$ and you have the pdf for your question!
Funnily enough after all that I'm not sure how to analytically find the mean and standard deviation of $f_{S_{t}}(x)$. But numerically it has mean $approx 7.244S_0$ and standard deviation $approx 2.176S_0$.
answered 11 hours ago
Erik Parkinson
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
This is actually a stochastic differential equation. See https://en.wikipedia.org/wiki/Geometric_Brownian_motion.
Say your bet starts at $S_0$, and we model it as $S_t$ over time. Then we can write
$$dS_{t}=mu S_{t},dt+sigma S_{t},dW_{t}$$
where $mu=.02$, $sigma=.03$. Here $W$ represents random brownian motion, so $dW$ is simply a draw from the standard normal. So letting each $dt = 1$, your interest rate is $dS_{t}=mu+sigma dW_{t}$. The analytic solution to this equation is
$$S_{t}=S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma W_{t}right)$$
The pdf of Brownian Motion $W_t$ is Normal with mean $0$ and variance $t$
$${displaystyle f_{W_{t}}(x)={frac {1}{sqrt {2pi t}}}e^{-x^{2}/(2t)}}$$
We want the pdf of $S_t$, $f_{S_{t}}(x)$. Because $S_{t} = g(W_{t})$ with
$$g(x) = S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma xright)$$
we can use the change of variable formula
$$f_{S_{t}}(x)=f_{W_{t}}(g^{-1}(x))|frac{dg^{-1}(x)}{dx}|$$
We have
$$g^{-1}(x) = frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma}$$
and
$$frac{dg^{-1}(x)}{dx} = frac{1}{sigma y}$$
Thus
$$f_{S_{t}}(x) = frac{1}{sigma y sqrt{2pi t}} exp{frac{-(frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma})^2}{2t}}$$
This is the general solution, plug in your values of $t=100$, $mu=.02$, and $sigma=.03$ and you have the pdf for your question!
Funnily enough after all that I'm not sure how to analytically find the mean and standard deviation of $f_{S_{t}}(x)$. But numerically it has mean $approx 7.244S_0$ and standard deviation $approx 2.176S_0$.
answered 11 hours ago
Erik Parkinson
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
This is actually a stochastic differential equation. See https://en.wikipedia.org/wiki/Geometric_Brownian_motion.
Say your bet starts at $S_0$, and we model it as $S_t$ over time. Then we can write
$$dS_{t}=mu S_{t},dt+sigma S_{t},dW_{t}$$
where $mu=.02$, $sigma=.03$. Here $W$ represents random brownian motion, so $dW$ is simply a draw from the standard normal. So letting each $dt = 1$, your interest rate is $dS_{t}=mu+sigma dW_{t}$. The analytic solution to this equation is
$$S_{t}=S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma W_{t}right)$$
The pdf of Brownian Motion $W_t$ is Normal with mean $0$ and variance $t$
$${displaystyle f_{W_{t}}(x)={frac {1}{sqrt {2pi t}}}e^{-x^{2}/(2t)}}$$
We want the pdf of $S_t$, $f_{S_{t}}(x)$. Because $S_{t} = g(W_{t})$ with
$$g(x) = S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma xright)$$
we can use the change of variable formula
$$f_{S_{t}}(x)=f_{W_{t}}(g^{-1}(x))|frac{dg^{-1}(x)}{dx}|$$
We have
$$g^{-1}(x) = frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma}$$
and
$$frac{dg^{-1}(x)}{dx} = frac{1}{sigma y}$$
Thus
$$f_{S_{t}}(x) = frac{1}{sigma y sqrt{2pi t}} exp{frac{-(frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma})^2}{2t}}$$
This is the general solution, plug in your values of $t=100$, $mu=.02$, and $sigma=.03$ and you have the pdf for your question!
Funnily enough after all that I'm not sure how to analytically find the mean and standard deviation of $f_{S_{t}}(x)$. But numerically it has mean $approx 7.244S_0$ and standard deviation $approx 2.176S_0$.
answered 11 hours ago
Erik Parkinson
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is actually a stochastic differential equation. See https://en.wikipedia.org/wiki/Geometric_Brownian_motion.
Say your bet starts at $S_0$, and we model it as $S_t$ over time. Then we can write
$$dS_{t}=mu S_{t},dt+sigma S_{t},dW_{t}$$
where $mu=.02$, $sigma=.03$. Here $W$ represents random brownian motion, so $dW$ is simply a draw from the standard normal. So letting each $dt = 1$, your interest rate is $dS_{t}=mu+sigma dW_{t}$. The analytic solution to this equation is
$$S_{t}=S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma W_{t}right)$$
The pdf of Brownian Motion $W_t$ is Normal with mean $0$ and variance $t$
$${displaystyle f_{W_{t}}(x)={frac {1}{sqrt {2pi t}}}e^{-x^{2}/(2t)}}$$
We want the pdf of $S_t$, $f_{S_{t}}(x)$. Because $S_{t} = g(W_{t})$ with
$$g(x) = S_{0}exp left(left(mu -{frac {sigma ^{2}}{2}}right)t+sigma xright)$$
we can use the change of variable formula
$$f_{S_{t}}(x)=f_{W_{t}}(g^{-1}(x))|frac{dg^{-1}(x)}{dx}|$$
We have
$$g^{-1}(x) = frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma}$$
and
$$frac{dg^{-1}(x)}{dx} = frac{1}{sigma y}$$
Thus
$$f_{S_{t}}(x) = frac{1}{sigma y sqrt{2pi t}} exp{frac{-(frac{ln{frac{y}{S_0}}-(mu-frac{sigma^2}{2})t}{sigma})^2}{2t}}$$
This is the general solution, plug in your values of $t=100$, $mu=.02$, and $sigma=.03$ and you have the pdf for your question!
Funnily enough after all that I'm not sure how to analytically find the mean and standard deviation of $f_{S_{t}}(x)$. But numerically it has mean $approx 7.244S_0$ and standard deviation $approx 2.176S_0$.
answered 11 hours ago
Erik Parkinson
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 11 hours ago
Erik Parkinson
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 11 hours ago
Erik Parkinson
6448
answered 11 hours ago
Erik Parkinson
6448
6448
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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Interesting question. Maybe you can take the logarithm of the product, which will turn it into a sum. Then you have a sum of random variables ... Anyway, just a thought.
– Matti P.
12 hours ago
I am considering running 10,000 simulations in matlab for 10 year predictions and just taking probabilities. I might be able to work out a confidence interval like that if I just find the expectation of 10,000 simulations. Hoping for a real analytical solution though.
– Frank
12 hours ago