Proof of Approximate / Exact Bayesian Computation
the ABC algorithm is given as
- Draw $theta sim pi(theta)$
- Simulate data $X sim pi(x | theta)$
- 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
asked Dec 3 at 13:24
charlie_wg
335
add a comment |
the ABC algorithm is given as
- Draw $theta sim pi(theta)$
- Simulate data $X sim pi(x | theta)$
- 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
asked Dec 3 at 13:24
charlie_wg
335
add a comment |
the ABC algorithm is given as
- Draw $theta sim pi(theta)$
- Simulate data $X sim pi(x | theta)$
- 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
asked Dec 3 at 13:24
charlie_wg
335
the ABC algorithm is given as
- Draw $theta sim pi(theta)$
- Simulate data $X sim pi(x | theta)$
- 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
asked Dec 3 at 13:24
charlie_wg
335
asked Dec 3 at 13:24
charlie_wg
335
asked Dec 3 at 13:24
charlie_wg
335
asked Dec 3 at 13:24
charlie_wg
335
asked Dec 3 at 13:24
charlie_wg
335
335
add a comment |
add a comment |
1 Answer
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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!
answered Dec 3 at 15:17
Xi'an
53.1k689343
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 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 at 18:28
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
votes
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!
answered Dec 3 at 15:17
Xi'an
53.1k689343
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 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 at 18:28
add a comment |
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!
answered Dec 3 at 15:17
Xi'an
53.1k689343
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 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 at 18:28
add a comment |
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!
answered Dec 3 at 15:17
Xi'an
53.1k689343
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!
answered Dec 3 at 15:17
Xi'an
53.1k689343
answered Dec 3 at 15:17
Xi'an
53.1k689343
answered Dec 3 at 15:17
Xi'an
53.1k689343
answered Dec 3 at 15:17
Xi'an
53.1k689343
53.1k689343
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 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 at 18:28
add a comment |
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 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 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 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 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 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 at 18:28
add a comment |
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