Is every probability distribution











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Say I have some arbitrary $N$-dimensional probability distribution $P$.



I'd like to create a deterministic function that accepts a vector of $N$ uniformly distributed numbers and generates an $N$ element vector where the elements of this vector are distributed according to $P$.



Is this always possible?










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    Say I have some arbitrary $N$-dimensional probability distribution $P$.



    I'd like to create a deterministic function that accepts a vector of $N$ uniformly distributed numbers and generates an $N$ element vector where the elements of this vector are distributed according to $P$.



    Is this always possible?










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Say I have some arbitrary $N$-dimensional probability distribution $P$.



      I'd like to create a deterministic function that accepts a vector of $N$ uniformly distributed numbers and generates an $N$ element vector where the elements of this vector are distributed according to $P$.



      Is this always possible?










      share|cite|improve this question













      Say I have some arbitrary $N$-dimensional probability distribution $P$.



      I'd like to create a deterministic function that accepts a vector of $N$ uniformly distributed numbers and generates an $N$ element vector where the elements of this vector are distributed according to $P$.



      Is this always possible?







      probability statistics






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      asked 2 days ago









      gigalord

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          1. If $N=1$ then it is always possible.
            Let $$ Xsim F, $$
            there exists $$F^{-1}(u)=inf {xin mathbb{R}colon F(x)ge u}$$ where $0< u<1$. Thus $$ F^{-1}(U)sim F$$
            with $Usim Unif(0,1)$. This method is called inverse transform.


          2. If $N>1$. Let $$ X=(X_1,X_2,dots,X_N)sim F $$
            where $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1,dots, X_Nle x_N).$$
            If $X_1,dots,X_N$ are independent, then $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1)cdots mathbb{P}(X_Nle x_N)=F_{X_1}(x_1)cdots F_{X_N}(x_N).$$
            Thus is possible to simulate each $X_i$ by inverse transform method and so
            $$ (F_{X_1}^{-1}(U_1),dots,F_{X_N}^{-1}(U_N))sim F$$
            where $U_1,dots,U_N$ i.i.d. with $U_1sim Unif(0,1)$.
            If $X_1,dots,X_N$ are dependent then a procedure that takes this into account is necessary.







          share|cite|improve this answer























          • So the sufficiency condition here is that $F$ is invertible?
            – gigalord
            yesterday










          • Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
            – Daniel Camarena Perez
            yesterday










          • In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
            – Ian
            yesterday












          • An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
            – Daniel Camarena Perez
            2 hours ago











          Your Answer





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          1. If $N=1$ then it is always possible.
            Let $$ Xsim F, $$
            there exists $$F^{-1}(u)=inf {xin mathbb{R}colon F(x)ge u}$$ where $0< u<1$. Thus $$ F^{-1}(U)sim F$$
            with $Usim Unif(0,1)$. This method is called inverse transform.


          2. If $N>1$. Let $$ X=(X_1,X_2,dots,X_N)sim F $$
            where $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1,dots, X_Nle x_N).$$
            If $X_1,dots,X_N$ are independent, then $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1)cdots mathbb{P}(X_Nle x_N)=F_{X_1}(x_1)cdots F_{X_N}(x_N).$$
            Thus is possible to simulate each $X_i$ by inverse transform method and so
            $$ (F_{X_1}^{-1}(U_1),dots,F_{X_N}^{-1}(U_N))sim F$$
            where $U_1,dots,U_N$ i.i.d. with $U_1sim Unif(0,1)$.
            If $X_1,dots,X_N$ are dependent then a procedure that takes this into account is necessary.







          share|cite|improve this answer























          • So the sufficiency condition here is that $F$ is invertible?
            – gigalord
            yesterday










          • Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
            – Daniel Camarena Perez
            yesterday










          • In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
            – Ian
            yesterday












          • An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
            – Daniel Camarena Perez
            2 hours ago















          up vote
          1
          down vote



          accepted











          1. If $N=1$ then it is always possible.
            Let $$ Xsim F, $$
            there exists $$F^{-1}(u)=inf {xin mathbb{R}colon F(x)ge u}$$ where $0< u<1$. Thus $$ F^{-1}(U)sim F$$
            with $Usim Unif(0,1)$. This method is called inverse transform.


          2. If $N>1$. Let $$ X=(X_1,X_2,dots,X_N)sim F $$
            where $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1,dots, X_Nle x_N).$$
            If $X_1,dots,X_N$ are independent, then $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1)cdots mathbb{P}(X_Nle x_N)=F_{X_1}(x_1)cdots F_{X_N}(x_N).$$
            Thus is possible to simulate each $X_i$ by inverse transform method and so
            $$ (F_{X_1}^{-1}(U_1),dots,F_{X_N}^{-1}(U_N))sim F$$
            where $U_1,dots,U_N$ i.i.d. with $U_1sim Unif(0,1)$.
            If $X_1,dots,X_N$ are dependent then a procedure that takes this into account is necessary.







          share|cite|improve this answer























          • So the sufficiency condition here is that $F$ is invertible?
            – gigalord
            yesterday










          • Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
            – Daniel Camarena Perez
            yesterday










          • In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
            – Ian
            yesterday












          • An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
            – Daniel Camarena Perez
            2 hours ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted







          1. If $N=1$ then it is always possible.
            Let $$ Xsim F, $$
            there exists $$F^{-1}(u)=inf {xin mathbb{R}colon F(x)ge u}$$ where $0< u<1$. Thus $$ F^{-1}(U)sim F$$
            with $Usim Unif(0,1)$. This method is called inverse transform.


          2. If $N>1$. Let $$ X=(X_1,X_2,dots,X_N)sim F $$
            where $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1,dots, X_Nle x_N).$$
            If $X_1,dots,X_N$ are independent, then $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1)cdots mathbb{P}(X_Nle x_N)=F_{X_1}(x_1)cdots F_{X_N}(x_N).$$
            Thus is possible to simulate each $X_i$ by inverse transform method and so
            $$ (F_{X_1}^{-1}(U_1),dots,F_{X_N}^{-1}(U_N))sim F$$
            where $U_1,dots,U_N$ i.i.d. with $U_1sim Unif(0,1)$.
            If $X_1,dots,X_N$ are dependent then a procedure that takes this into account is necessary.







          share|cite|improve this answer















          1. If $N=1$ then it is always possible.
            Let $$ Xsim F, $$
            there exists $$F^{-1}(u)=inf {xin mathbb{R}colon F(x)ge u}$$ where $0< u<1$. Thus $$ F^{-1}(U)sim F$$
            with $Usim Unif(0,1)$. This method is called inverse transform.


          2. If $N>1$. Let $$ X=(X_1,X_2,dots,X_N)sim F $$
            where $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1,dots, X_Nle x_N).$$
            If $X_1,dots,X_N$ are independent, then $$F(x_1,dots,x_N)=mathbb{P}(X_1le x_1)cdots mathbb{P}(X_Nle x_N)=F_{X_1}(x_1)cdots F_{X_N}(x_N).$$
            Thus is possible to simulate each $X_i$ by inverse transform method and so
            $$ (F_{X_1}^{-1}(U_1),dots,F_{X_N}^{-1}(U_N))sim F$$
            where $U_1,dots,U_N$ i.i.d. with $U_1sim Unif(0,1)$.
            If $X_1,dots,X_N$ are dependent then a procedure that takes this into account is necessary.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered 2 days ago









          Daniel Camarena Perez

          56228




          56228












          • So the sufficiency condition here is that $F$ is invertible?
            – gigalord
            yesterday










          • Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
            – Daniel Camarena Perez
            yesterday










          • In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
            – Ian
            yesterday












          • An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
            – Daniel Camarena Perez
            2 hours ago


















          • So the sufficiency condition here is that $F$ is invertible?
            – gigalord
            yesterday










          • Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
            – Daniel Camarena Perez
            yesterday










          • In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
            – Ian
            yesterday












          • An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
            – Daniel Camarena Perez
            2 hours ago
















          So the sufficiency condition here is that $F$ is invertible?
          – gigalord
          yesterday




          So the sufficiency condition here is that $F$ is invertible?
          – gigalord
          yesterday












          Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
          – Daniel Camarena Perez
          yesterday




          Yes, although this is a very strong hypothesis, and in practice difficult to achieve. Also, note that when $N = 1$ we use a generalized inverse, and even this method is not widely used in practice. In practice, Markov Chain Monte Carlo is used to simulate a complex distribution.
          – Daniel Camarena Perez
          yesterday












          In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
          – Ian
          yesterday






          In the one-dimensional case the "probability integral transform" is used quite routinely. It's especially easy to implement in the discrete setting, where there are also alternatives that avoid the need for a binary search lookup (by using some preprocessing).
          – Ian
          yesterday














          An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
          – Daniel Camarena Perez
          2 hours ago




          An example with dependence is a Markov chain columbia.edu/~ks20/4703-Sigman/4703-07-Notes-MC.pdf
          – Daniel Camarena Perez
          2 hours ago


















           

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