compound interest when interest is a random variable












1














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







share|cite|improve this question






















  • 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.
    Jan 4 at 9:30










  • 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
    Jan 4 at 9:45


















1














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







share|cite|improve this question






















  • 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.
    Jan 4 at 9:30










  • 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
    Jan 4 at 9:45
















1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 9:25









FrankFrank

15710




15710












  • 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.
    Jan 4 at 9:30










  • 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
    Jan 4 at 9:45




















  • 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.
    Jan 4 at 9:30










  • 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
    Jan 4 at 9:45


















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.
Jan 4 at 9:30




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.
Jan 4 at 9:30












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
Jan 4 at 9:45






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
Jan 4 at 9:45












1 Answer
1






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oldest

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0














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 $approx 2.176S_0$.






share|cite|improve this answer








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    0














    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 $approx 2.176S_0$.






    share|cite|improve this answer








    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.























      0














      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 $approx 2.176S_0$.






      share|cite|improve this answer








      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.





















        0












        0








        0






        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 $approx 2.176S_0$.






        share|cite|improve this answer








        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 $approx 2.176S_0$.







        share|cite|improve this answer








        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.









        share|cite|improve this answer



        share|cite|improve this answer






        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 Jan 4 at 11:02









        Erik ParkinsonErik Parkinson

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        9159




        New contributor




        Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        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.






























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