Variational approximation for importance sampling

Abstract

We propose an importance sampling algorithm with proposal distribution obtained from variational approximation. This method combines the strength of both importance sampling and variational method. On one hand, this method avoids the bias from variational method. On the other hand, variational approximation provides a way to design the proposal distribution for the importance sampling algorithm. Theoretical justification of the proposed method is provided. Numerical results show that using variational approximation as the proposal can improve the performance of importance sampling and sequential importance sampling.

Appendices

Appendices

Proof of Lemma 1

Proof

We have $\lim _{n\rightarrow \infty } \beta _{2,n} = 1$ immediately from the definition of convergence in (4).

Now we prove $\lim _{n\rightarrow \infty } \beta _{1,n} = 1$. For $\forall \ \epsilon > 0$ and $\delta > 0$, define $I_1^{(n)} = \{x:\frac{p_n(x)}{q(x)}< 1-\epsilon \}$, $I_2^{(n)} = \{x:1-\epsilon \le \frac{p_n(x)}{q(x)}< 1+\delta \}$, and $I_3^{(n)} = \{x:\frac{p_n(x)}{q(x)}\ge 1+\delta \}$.

From (4), we have for any given $\epsilon >0$, there exists $N\in \mathbb N$ such that for all $n>N$, we have

$$\begin{aligned} \text {ess}\inf \,\frac{p_n}{q} > 1-\epsilon .\end{aligned}$$

By the definition of essential infimum, we have

$$\begin{aligned} \sup \{b \in \mathbb R: \mu (\{x: p_n(x)/q(x) < b \}) = 0 \} > 1-\epsilon , \end{aligned}$$

which implies

$$\begin{aligned} \mu (I_1^{(n)})=\mu (\{x: p_n(x)/q(x) < 1-\epsilon \}) = 0. \end{aligned}$$

Then we have

$$\begin{aligned} \int _{I_1^{(n)}}\frac{p_n}{q}\,q\,dx = \int _{I_1^{(n)}}p_n\,dx =0 \ \ \ \text{ for } n > N.\end{aligned}$$

So

$$\begin{aligned} 1 = \int _{\mathbb {R}} p_n \,dx= & {} \int _{I_1^{(n)}}\frac{p_n}{q}\,q\,dx + \int _{I_2^{(n)}}\frac{p_n}{q}\,q\,dx + \int _{I_3^{(n)}}\frac{p_n}{q}\,q\,dx \\= & {} \int _{I_2^{(n)}}\frac{p_n}{q}\,q\,dx + \int _{I_3^{(n)}}\frac{p_n}{q}\,q\,dx \ \ \ \ \ \text{ for } n > N. \end{aligned}$$

From the definitions of $I_2^{(n)}$ and $I_3^{(n)}$, we have

$$\begin{aligned} 1= & {} \int _{I_2^{(n)}}\frac{p_n}{q}\,q\,dx + \int _{I_3^{(n)}}\frac{p_n}{q}\,q\,dx \nonumber \\\ge & {} (1-\epsilon )\int _{I_2^{(n)}}\,q\,dx + (1+\delta )\int _{I_3^{(n)}}\,q\,dx \ \ \ \ \text{ for } n > N. \end{aligned}$$

(10)

Similarly, we also have

$$\begin{aligned} 1=\int _{\mathbb {R}} q \,dx= & {} \int _{I_1^{(n)}}q\,dx + \int _{I_2^{(n)}}q\,dx + \int _{I_3^{(n)}}q\,dx \nonumber \\= & {} \int _{I_2^{(n)}}q\,dx + \int _{I_3^{(n)}}q\,dx \ \ \ \ \text{ for } n > N. \end{aligned}$$

(11)

From (10) and (11), the following inequality holds:

$$\begin{aligned} 1\ge & {} (1-\epsilon )\int _{I_2^{(n)}}\,q\,dx + (1+\delta )\left( 1-\int _{I_2^{(n)}}\,q\,dx\right) \nonumber \\= & {} (1+\delta ) - (\epsilon +\delta ) \int _{I_2^{(n)}}\,q\,dx \ \ \ \ \text{ for } n > N. \end{aligned}$$

(12)

Suppose $\limsup \limits _{n \rightarrow \infty }\int _{I_2^{(n)}}\,q\,dx = \theta (\epsilon , \delta ) \in [0,1]$, then $\liminf \limits _{n \rightarrow \infty }\int _{I_3^{(n)}}\,q\,dx = 1-\theta (\epsilon , \delta )$ based on (11). Since the definition of $I_3^{(n)}$ depends only on $\delta$, not $\epsilon$, we know that $\liminf \limits _{n \rightarrow \infty }\int _{I_3^{(n)}}\,q\,dx$ also depends only on $\delta$, not $\epsilon$. Thus $\theta (\epsilon , \delta ) = \theta (\delta )$ does not depend on $\epsilon$.

Taking limit inferior on both sides of (12), we have

$$\begin{aligned} 1\ge & {} (1+\delta ) -\limsup \limits _{n \rightarrow \infty }\left( (\epsilon +\delta ) \int _{I_2^{(n)}}\,q\,dx \right) = (1+\delta ) - (\epsilon +\delta )\theta (\delta ). \end{aligned}$$

(13)

Therefore,

$$\begin{aligned} \theta (\delta ) \ge \frac{\delta }{\delta +\epsilon }. \end{aligned}$$

(14)

Note that (14) is true for any $\epsilon > 0$ and $\delta > 0$ selected at the beginning of the proof. Since the left hand side of (14) does not depend on $\epsilon$, letting $\epsilon \rightarrow 0$ on the right hand side of (14), we have $\theta (\delta ) \ge 1$. On the other hand, $\limsup \limits _{n \rightarrow \infty }\int _{I_2^{(n)}}\,q\,dx = \theta (\delta ) \in [0,1]$. Therefore we have $\theta (\delta )=1$, which implies $\liminf \limits _{n \rightarrow \infty }\int _{I_3^{(n)}}\,q\,dx =1-\theta (\delta )= 0$ for any $\delta >0$. From the definition of $\beta _{1,n}$, we have $\lim _{n\rightarrow \infty } \beta _{1,n} = 1$.

Since $\mu (\{x: p_n(x)/q(x) < \beta _{2,n} \}) = \mu (\{x: p_n(x)/q(x) > \beta _{1,n}^{-1} \}) =0$, we have

$$\begin{aligned} D_f(p_n||q) = \int _{\{\beta _{2,n}\le \frac{p_n}{q}\le \beta _{1,n}^{-1}\}} f\left( \frac{p_n}{q}\right) \,q\,dx \le \sup _{\beta _{2,n}\le \beta \le \beta _{1,n}^{-1}}|f(\beta )|.\end{aligned}$$

Letting $n\rightarrow \infty$, due to the continuity of f at 1, we have $\lim _{n\rightarrow \infty } D_f(p_n||q) \le f(1) = 0$. $\square$

Proof of Theorem 1

Proof

From Lemma 1, we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \beta _{1,n} =\lim _{n\rightarrow \infty } \beta _{2,n}= 1. \end{aligned}$$

By L’Hospital’s rule, we have $\lim _{t \rightarrow 1}\kappa (t) = 1$, where $\kappa (t)$ is defined in (5). Therefore, take limit on the both sides of (6) and (7), we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{KL(p_n||q)}{KL(q||p_n)} = 1 \ , \ \ \lim _{n\rightarrow \infty } \frac{KL(p_n||q)}{\chi ^2(p_n||q)} = \frac{1}{2} .\end{aligned}$$

$\square$