Sample from aggregate portfolio distribution versus individual asset distributions











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Suppose I have three assets $x_1,x_2,x_3$ in a portfolio with weights $W=begin{bmatrix} w_1 \ w_2 \ w_3 end{bmatrix} $, expected returns $R=begin{bmatrix} mu_1 \ mu_2 \ mu_3 end{bmatrix}$, and a covariance matrix $V$.



The expected return of my portfolio is $mu_p=W^TR$ and the variance of my portfolio is $sigma^2_p=W^TVW$.



I would like to run Monte Carlo simulations on my portfolio using a normal distribution.



I can do this either by:




  1. Sampling from the distribution of portfolio returns $N(mu_p,sigma^2_p)$.

  2. Sample from the three individual asset returns and use those three returns to compute my overall portfolio return.


First, how would I accomplish the second approach (would I be sampling from a multivariate normal distribution)?



Second, are these two approaches equivalent as long as I assume that the weights $W$ of my portfolio remain the same?










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    Suppose I have three assets $x_1,x_2,x_3$ in a portfolio with weights $W=begin{bmatrix} w_1 \ w_2 \ w_3 end{bmatrix} $, expected returns $R=begin{bmatrix} mu_1 \ mu_2 \ mu_3 end{bmatrix}$, and a covariance matrix $V$.



    The expected return of my portfolio is $mu_p=W^TR$ and the variance of my portfolio is $sigma^2_p=W^TVW$.



    I would like to run Monte Carlo simulations on my portfolio using a normal distribution.



    I can do this either by:




    1. Sampling from the distribution of portfolio returns $N(mu_p,sigma^2_p)$.

    2. Sample from the three individual asset returns and use those three returns to compute my overall portfolio return.


    First, how would I accomplish the second approach (would I be sampling from a multivariate normal distribution)?



    Second, are these two approaches equivalent as long as I assume that the weights $W$ of my portfolio remain the same?










    share|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Suppose I have three assets $x_1,x_2,x_3$ in a portfolio with weights $W=begin{bmatrix} w_1 \ w_2 \ w_3 end{bmatrix} $, expected returns $R=begin{bmatrix} mu_1 \ mu_2 \ mu_3 end{bmatrix}$, and a covariance matrix $V$.



      The expected return of my portfolio is $mu_p=W^TR$ and the variance of my portfolio is $sigma^2_p=W^TVW$.



      I would like to run Monte Carlo simulations on my portfolio using a normal distribution.



      I can do this either by:




      1. Sampling from the distribution of portfolio returns $N(mu_p,sigma^2_p)$.

      2. Sample from the three individual asset returns and use those three returns to compute my overall portfolio return.


      First, how would I accomplish the second approach (would I be sampling from a multivariate normal distribution)?



      Second, are these two approaches equivalent as long as I assume that the weights $W$ of my portfolio remain the same?










      share|improve this question













      Suppose I have three assets $x_1,x_2,x_3$ in a portfolio with weights $W=begin{bmatrix} w_1 \ w_2 \ w_3 end{bmatrix} $, expected returns $R=begin{bmatrix} mu_1 \ mu_2 \ mu_3 end{bmatrix}$, and a covariance matrix $V$.



      The expected return of my portfolio is $mu_p=W^TR$ and the variance of my portfolio is $sigma^2_p=W^TVW$.



      I would like to run Monte Carlo simulations on my portfolio using a normal distribution.



      I can do this either by:




      1. Sampling from the distribution of portfolio returns $N(mu_p,sigma^2_p)$.

      2. Sample from the three individual asset returns and use those three returns to compute my overall portfolio return.


      First, how would I accomplish the second approach (would I be sampling from a multivariate normal distribution)?



      Second, are these two approaches equivalent as long as I assume that the weights $W$ of my portfolio remain the same?







      monte-carlo statistics modern-portfolio-theory






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      asked Nov 23 at 14:31









      cpage

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          2 Answers
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          accepted










          For the first case, you would directly sample $n$ random normals $x$ and compute:
          $$R^p_i = mu_p + sigma_p x_i, i in [1,n]$$



          For the second case, you can sample $n$ x $3$ independent normals, compute the Cholesky decomposition matrix $C$ of $V$, which is the matrix $C$ such that $V=C^t C$, and get $n$ samples of vectors $X$ of size 3.



          The return $R_i$ for random draw $i$ is given by:
          $$R_i = mu_p + C . X_i, i in [1,n]$$
          You can check for high values of $n$ the convergence towards the limit values:
          $$E(R_i) = R$$
          $$Cov(R_i) = V$$
          The portfolio return is then computed as:
          $$R^p_i = W.T R_i$$
          and you can check it converges towards the same mean and variance $mu_p$, $sigma_p^2$ for a large enough $n$.



          The two approaches are mathematically equivalent as a linear combination of independent normals is normally distributed. This works so long as the random normal variables generated are iid gaussian normals.



          With numpy, iid normals can be generated with np.random.normal. As pointed out below, np.random.multivariate_normal can be used to generate the multivariate gaussian vector.






          share|improve this answer



















          • 1




            How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
            – cpage
            Nov 23 at 16:16












          • good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
            – Sebapi
            Nov 24 at 9:52




















          up vote
          0
          down vote













          Basically what @sebapi said.
          "The two approaches are equivalent so long as the random normal variables generated are iid gaussian normals."



          Q: How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?



          A: You might use scipy.stats.multivariate_normal (rv = multivariate_normal(mean=None, cov=1, allow_singular=False))






          share|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            For the first case, you would directly sample $n$ random normals $x$ and compute:
            $$R^p_i = mu_p + sigma_p x_i, i in [1,n]$$



            For the second case, you can sample $n$ x $3$ independent normals, compute the Cholesky decomposition matrix $C$ of $V$, which is the matrix $C$ such that $V=C^t C$, and get $n$ samples of vectors $X$ of size 3.



            The return $R_i$ for random draw $i$ is given by:
            $$R_i = mu_p + C . X_i, i in [1,n]$$
            You can check for high values of $n$ the convergence towards the limit values:
            $$E(R_i) = R$$
            $$Cov(R_i) = V$$
            The portfolio return is then computed as:
            $$R^p_i = W.T R_i$$
            and you can check it converges towards the same mean and variance $mu_p$, $sigma_p^2$ for a large enough $n$.



            The two approaches are mathematically equivalent as a linear combination of independent normals is normally distributed. This works so long as the random normal variables generated are iid gaussian normals.



            With numpy, iid normals can be generated with np.random.normal. As pointed out below, np.random.multivariate_normal can be used to generate the multivariate gaussian vector.






            share|improve this answer



















            • 1




              How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
              – cpage
              Nov 23 at 16:16












            • good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
              – Sebapi
              Nov 24 at 9:52

















            up vote
            2
            down vote



            accepted










            For the first case, you would directly sample $n$ random normals $x$ and compute:
            $$R^p_i = mu_p + sigma_p x_i, i in [1,n]$$



            For the second case, you can sample $n$ x $3$ independent normals, compute the Cholesky decomposition matrix $C$ of $V$, which is the matrix $C$ such that $V=C^t C$, and get $n$ samples of vectors $X$ of size 3.



            The return $R_i$ for random draw $i$ is given by:
            $$R_i = mu_p + C . X_i, i in [1,n]$$
            You can check for high values of $n$ the convergence towards the limit values:
            $$E(R_i) = R$$
            $$Cov(R_i) = V$$
            The portfolio return is then computed as:
            $$R^p_i = W.T R_i$$
            and you can check it converges towards the same mean and variance $mu_p$, $sigma_p^2$ for a large enough $n$.



            The two approaches are mathematically equivalent as a linear combination of independent normals is normally distributed. This works so long as the random normal variables generated are iid gaussian normals.



            With numpy, iid normals can be generated with np.random.normal. As pointed out below, np.random.multivariate_normal can be used to generate the multivariate gaussian vector.






            share|improve this answer



















            • 1




              How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
              – cpage
              Nov 23 at 16:16












            • good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
              – Sebapi
              Nov 24 at 9:52















            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            For the first case, you would directly sample $n$ random normals $x$ and compute:
            $$R^p_i = mu_p + sigma_p x_i, i in [1,n]$$



            For the second case, you can sample $n$ x $3$ independent normals, compute the Cholesky decomposition matrix $C$ of $V$, which is the matrix $C$ such that $V=C^t C$, and get $n$ samples of vectors $X$ of size 3.



            The return $R_i$ for random draw $i$ is given by:
            $$R_i = mu_p + C . X_i, i in [1,n]$$
            You can check for high values of $n$ the convergence towards the limit values:
            $$E(R_i) = R$$
            $$Cov(R_i) = V$$
            The portfolio return is then computed as:
            $$R^p_i = W.T R_i$$
            and you can check it converges towards the same mean and variance $mu_p$, $sigma_p^2$ for a large enough $n$.



            The two approaches are mathematically equivalent as a linear combination of independent normals is normally distributed. This works so long as the random normal variables generated are iid gaussian normals.



            With numpy, iid normals can be generated with np.random.normal. As pointed out below, np.random.multivariate_normal can be used to generate the multivariate gaussian vector.






            share|improve this answer














            For the first case, you would directly sample $n$ random normals $x$ and compute:
            $$R^p_i = mu_p + sigma_p x_i, i in [1,n]$$



            For the second case, you can sample $n$ x $3$ independent normals, compute the Cholesky decomposition matrix $C$ of $V$, which is the matrix $C$ such that $V=C^t C$, and get $n$ samples of vectors $X$ of size 3.



            The return $R_i$ for random draw $i$ is given by:
            $$R_i = mu_p + C . X_i, i in [1,n]$$
            You can check for high values of $n$ the convergence towards the limit values:
            $$E(R_i) = R$$
            $$Cov(R_i) = V$$
            The portfolio return is then computed as:
            $$R^p_i = W.T R_i$$
            and you can check it converges towards the same mean and variance $mu_p$, $sigma_p^2$ for a large enough $n$.



            The two approaches are mathematically equivalent as a linear combination of independent normals is normally distributed. This works so long as the random normal variables generated are iid gaussian normals.



            With numpy, iid normals can be generated with np.random.normal. As pointed out below, np.random.multivariate_normal can be used to generate the multivariate gaussian vector.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Nov 24 at 9:46

























            answered Nov 23 at 15:09









            Sebapi

            865




            865








            • 1




              How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
              – cpage
              Nov 23 at 16:16












            • good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
              – Sebapi
              Nov 24 at 9:52
















            • 1




              How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
              – cpage
              Nov 23 at 16:16












            • good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
              – Sebapi
              Nov 24 at 9:52










            1




            1




            How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
            – cpage
            Nov 23 at 16:16






            How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?
            – cpage
            Nov 23 at 16:16














            good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
            – Sebapi
            Nov 24 at 9:52






            good question from @cpage on a 5x5 historical covariance matrix, I get 0.27s for 1e6 samples from np.random.multivariate_normal and 5.3s using np.random.normal with a loop in python. The samples have similar statistics but different values.
            – Sebapi
            Nov 24 at 9:52












            up vote
            0
            down vote













            Basically what @sebapi said.
            "The two approaches are equivalent so long as the random normal variables generated are iid gaussian normals."



            Q: How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?



            A: You might use scipy.stats.multivariate_normal (rv = multivariate_normal(mean=None, cov=1, allow_singular=False))






            share|improve this answer

























              up vote
              0
              down vote













              Basically what @sebapi said.
              "The two approaches are equivalent so long as the random normal variables generated are iid gaussian normals."



              Q: How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?



              A: You might use scipy.stats.multivariate_normal (rv = multivariate_normal(mean=None, cov=1, allow_singular=False))






              share|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                Basically what @sebapi said.
                "The two approaches are equivalent so long as the random normal variables generated are iid gaussian normals."



                Q: How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?



                A: You might use scipy.stats.multivariate_normal (rv = multivariate_normal(mean=None, cov=1, allow_singular=False))






                share|improve this answer












                Basically what @sebapi said.
                "The two approaches are equivalent so long as the random normal variables generated are iid gaussian normals."



                Q: How does this compare to using docs.scipy.org/doc/numpy-1.15.1/reference/generated/…?



                A: You might use scipy.stats.multivariate_normal (rv = multivariate_normal(mean=None, cov=1, allow_singular=False))







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 23 at 18:32









                TomDecimus

                917




                917






























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