Proof of Approximate / Exact Bayesian Computation












5














the ABC algorithm is given as




  1. Draw $theta sim pi(theta)$

  2. Simulate data $X sim pi(x | theta)$

  3. Accept $theta$ if $rho(X, D) < varepsilon$


where $pi(theta)$ is the prior, $pi(x | theta)$ is the likelihood, $rho(cdot | cdot)$ is some distance measure, $D$ is the observed data and $varepsilon$ is the tolerance that represents a trade off between accuracy and computability.



Generally, in papers that I have seen on this, a proof is given where it states we actually sample from $pi_{varepsilon} = pi(theta | rho(X, D) < varepsilon)$ and then if $varepsilon to 0$, this converges to the true posterior $pi(theta | D)$.



If in Step 3, we had



3*. Accept $theta$ if $X = D$



I was wondering if anyone knew how to prove that in this new algorithm, we sample from the true posterior? So there is no $varepsilon to 0$ argument?



Thanks in advance to anyone that can offer some help!






bayesian abc







share|cite|improve this question



























    5














    the ABC algorithm is given as




    1. Draw $theta sim pi(theta)$

    2. Simulate data $X sim pi(x | theta)$

    3. Accept $theta$ if $rho(X, D) < varepsilon$


    where $pi(theta)$ is the prior, $pi(x | theta)$ is the likelihood, $rho(cdot | cdot)$ is some distance measure, $D$ is the observed data and $varepsilon$ is the tolerance that represents a trade off between accuracy and computability.



    Generally, in papers that I have seen on this, a proof is given where it states we actually sample from $pi_{varepsilon} = pi(theta | rho(X, D) < varepsilon)$ and then if $varepsilon to 0$, this converges to the true posterior $pi(theta | D)$.



    If in Step 3, we had



    3*. Accept $theta$ if $X = D$



    I was wondering if anyone knew how to prove that in this new algorithm, we sample from the true posterior? So there is no $varepsilon to 0$ argument?



    Thanks in advance to anyone that can offer some help!






    bayesian abc







    share|cite|improve this question

























      5












      5








      5


      2





      the ABC algorithm is given as




      1. Draw $theta sim pi(theta)$

      2. Simulate data $X sim pi(x | theta)$

      3. Accept $theta$ if $rho(X, D) < varepsilon$


      where $pi(theta)$ is the prior, $pi(x | theta)$ is the likelihood, $rho(cdot | cdot)$ is some distance measure, $D$ is the observed data and $varepsilon$ is the tolerance that represents a trade off between accuracy and computability.



      Generally, in papers that I have seen on this, a proof is given where it states we actually sample from $pi_{varepsilon} = pi(theta | rho(X, D) < varepsilon)$ and then if $varepsilon to 0$, this converges to the true posterior $pi(theta | D)$.



      If in Step 3, we had



      3*. Accept $theta$ if $X = D$



      I was wondering if anyone knew how to prove that in this new algorithm, we sample from the true posterior? So there is no $varepsilon to 0$ argument?



      Thanks in advance to anyone that can offer some help!






      bayesian abc







      share|cite|improve this question













      the ABC algorithm is given as




      1. Draw $theta sim pi(theta)$

      2. Simulate data $X sim pi(x | theta)$

      3. Accept $theta$ if $rho(X, D) < varepsilon$


      where $pi(theta)$ is the prior, $pi(x | theta)$ is the likelihood, $rho(cdot | cdot)$ is some distance measure, $D$ is the observed data and $varepsilon$ is the tolerance that represents a trade off between accuracy and computability.



      Generally, in papers that I have seen on this, a proof is given where it states we actually sample from $pi_{varepsilon} = pi(theta | rho(X, D) < varepsilon)$ and then if $varepsilon to 0$, this converges to the true posterior $pi(theta | D)$.



      If in Step 3, we had



      3*. Accept $theta$ if $X = D$



      I was wondering if anyone knew how to prove that in this new algorithm, we sample from the true posterior? So there is no $varepsilon to 0$ argument?



      Thanks in advance to anyone that can offer some help!






      bayesian abc




      bayesian abc






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 3 '18 at 13:24









      charlie_wg

      335




      335






















          1 Answer
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          active

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          6














          This case is the original version of the algorithm, as in Rubin (1984) and Tavaré et al. (1997). Assuming that $$mathbb{P}_theta(X=D)>0$$ the values of $theta$ that come out of the algorithm are distributed from a distribution with density proportional to
          $$pi(theta) times mathbb{P}_theta(X=D)$$
          since the algorithm generates the pair $(theta,mathbb{I}_{X=D})$ with joint distribution
          $$pi(theta) times mathbb{P}_theta(X=D)^{mathbb{I}_{X=D}} times
          mathbb{P}_theta(Xne D)^{mathbb{I}_{Xne D}}$$
          Conditioning on $mathbb{I}_{X=D}=1$ leads to
          $$theta|mathbb{I}_{X=D}=1 sim pi(theta) times mathbb{P}_theta(X=D)Big/int pi(theta) times mathbb{P}_theta(X=D) ,text{d}theta$$
          which is the posterior distribution.



          On the side, I gave this very proof in class a few hours ago!






          share|cite|improve this answer

















          • 1




            Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
            – charlie_wg
            Dec 3 '18 at 16:04






          • 1




            Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
            – Xi'an
            Dec 3 '18 at 18:28











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6














          This case is the original version of the algorithm, as in Rubin (1984) and Tavaré et al. (1997). Assuming that $$mathbb{P}_theta(X=D)>0$$ the values of $theta$ that come out of the algorithm are distributed from a distribution with density proportional to
          $$pi(theta) times mathbb{P}_theta(X=D)$$
          since the algorithm generates the pair $(theta,mathbb{I}_{X=D})$ with joint distribution
          $$pi(theta) times mathbb{P}_theta(X=D)^{mathbb{I}_{X=D}} times
          mathbb{P}_theta(Xne D)^{mathbb{I}_{Xne D}}$$
          Conditioning on $mathbb{I}_{X=D}=1$ leads to
          $$theta|mathbb{I}_{X=D}=1 sim pi(theta) times mathbb{P}_theta(X=D)Big/int pi(theta) times mathbb{P}_theta(X=D) ,text{d}theta$$
          which is the posterior distribution.



          On the side, I gave this very proof in class a few hours ago!






          share|cite|improve this answer

















          • 1




            Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
            – charlie_wg
            Dec 3 '18 at 16:04






          • 1




            Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
            – Xi'an
            Dec 3 '18 at 18:28
















          6














          This case is the original version of the algorithm, as in Rubin (1984) and Tavaré et al. (1997). Assuming that $$mathbb{P}_theta(X=D)>0$$ the values of $theta$ that come out of the algorithm are distributed from a distribution with density proportional to
          $$pi(theta) times mathbb{P}_theta(X=D)$$
          since the algorithm generates the pair $(theta,mathbb{I}_{X=D})$ with joint distribution
          $$pi(theta) times mathbb{P}_theta(X=D)^{mathbb{I}_{X=D}} times
          mathbb{P}_theta(Xne D)^{mathbb{I}_{Xne D}}$$
          Conditioning on $mathbb{I}_{X=D}=1$ leads to
          $$theta|mathbb{I}_{X=D}=1 sim pi(theta) times mathbb{P}_theta(X=D)Big/int pi(theta) times mathbb{P}_theta(X=D) ,text{d}theta$$
          which is the posterior distribution.



          On the side, I gave this very proof in class a few hours ago!






          share|cite|improve this answer

















          • 1




            Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
            – charlie_wg
            Dec 3 '18 at 16:04






          • 1




            Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
            – Xi'an
            Dec 3 '18 at 18:28














          6












          6








          6






          This case is the original version of the algorithm, as in Rubin (1984) and Tavaré et al. (1997). Assuming that $$mathbb{P}_theta(X=D)>0$$ the values of $theta$ that come out of the algorithm are distributed from a distribution with density proportional to
          $$pi(theta) times mathbb{P}_theta(X=D)$$
          since the algorithm generates the pair $(theta,mathbb{I}_{X=D})$ with joint distribution
          $$pi(theta) times mathbb{P}_theta(X=D)^{mathbb{I}_{X=D}} times
          mathbb{P}_theta(Xne D)^{mathbb{I}_{Xne D}}$$
          Conditioning on $mathbb{I}_{X=D}=1$ leads to
          $$theta|mathbb{I}_{X=D}=1 sim pi(theta) times mathbb{P}_theta(X=D)Big/int pi(theta) times mathbb{P}_theta(X=D) ,text{d}theta$$
          which is the posterior distribution.



          On the side, I gave this very proof in class a few hours ago!






          share|cite|improve this answer












          This case is the original version of the algorithm, as in Rubin (1984) and Tavaré et al. (1997). Assuming that $$mathbb{P}_theta(X=D)>0$$ the values of $theta$ that come out of the algorithm are distributed from a distribution with density proportional to
          $$pi(theta) times mathbb{P}_theta(X=D)$$
          since the algorithm generates the pair $(theta,mathbb{I}_{X=D})$ with joint distribution
          $$pi(theta) times mathbb{P}_theta(X=D)^{mathbb{I}_{X=D}} times
          mathbb{P}_theta(Xne D)^{mathbb{I}_{Xne D}}$$
          Conditioning on $mathbb{I}_{X=D}=1$ leads to
          $$theta|mathbb{I}_{X=D}=1 sim pi(theta) times mathbb{P}_theta(X=D)Big/int pi(theta) times mathbb{P}_theta(X=D) ,text{d}theta$$
          which is the posterior distribution.



          On the side, I gave this very proof in class a few hours ago!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 15:17









          Xi'an

          53.5k689346




          53.5k689346








          • 1




            Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
            – charlie_wg
            Dec 3 '18 at 16:04






          • 1




            Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
            – Xi'an
            Dec 3 '18 at 18:28














          • 1




            Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
            – charlie_wg
            Dec 3 '18 at 16:04






          • 1




            Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
            – Xi'an
            Dec 3 '18 at 18:28








          1




          1




          Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
          – charlie_wg
          Dec 3 '18 at 16:04




          Thanks Xi'an! If this was extended outside of the Bayesian context, so for arbitrary functions $f_{1}(x)$ and $f_{2}(x)$, does this work to find samples from $f propto f_{1}f_{2}$?
          – charlie_wg
          Dec 3 '18 at 16:04




          1




          1




          Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
          – Xi'an
          Dec 3 '18 at 18:28




          Yes, indeed!, this is actually the most rudimentary form of acceptance-rejection: simulate $X$ from $f_1$ and accept if the realisation of $Ysim f_2$ is equal to the realisation of $X$.
          – Xi'an
          Dec 3 '18 at 18:28


















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