# Application of the law of large numbers

In the book by Christian Robert and George Casella on Monte Carlo Statistical Methods, they use an argument of LLN on pages 551 and 552. I'm attaching the argument in this screen shot

$t$ is representing time. $i$ is representing the Monte Carlo sample so that

$$z_t^{(i)} \sim g \quad\quad \text{is the ith sample of z at time t}.$$

I understand how the first expression is obtained, so take is as given. What i don't follow is the argument after this first expression. the text reads "using the LLN for the sum within the exponential", but with respect to what index? isn't (i) fixed? It seems that the expectation just popped up inside the exponential (with respect to $g$).

What I expect from this part is an argument explaining the degeneracy of the weight $w_t^{(i)}$ that have tendency to go to zero for most of the $i$'s as $t$ increases (which I think happens for small values of $i$, since this method is consistent, and letting $i\to \infty$ we get convergence)

EDIT:

It is clear for me now that when we assume that the pdfs are time independent and we have $z_t$ as i.i.d. over $t$, we will be able to apply the LLN inside the exponential by writing $$\exp\left\{-t \sum_{\ell = 1}^t \frac{1}{t} \log\left( g(z_\ell^{(i)})/f(z_\ell^{(i)} \right) \right\}$$

But what happens when we try to take the limit ? We have to do this inside the exponential and only to a part of the argument, namely $\sum_{\ell = 1}^t \frac{1}{t} \log\left( g(z_\ell^{(i)})/f(z_\ell^{(i)} \right)$. So what happens to that $-t$ outside the limit? and what can you imply now on $w_t^{(i)}$ on the left hand side of the proportionality relation?

• The lln seems to be applied on index lowercase L. Saying sum(X_l)=t*1/tsum(X_l)->tE(X). I don't understand your problem with the degeneracy - you seem to re-tell it correctly so I don't think there is a problem – svendvn Feb 10 '17 at 8:43
• $X_l$ is a Markov process, and not a sequence of iid variables. I also do not see how can you take the limit of a part of the expression. What does this implies on the left hand side? Are you taking limit also for $w_t^{(i)}$. – user144410 Feb 10 '17 at 13:11
• So, I see now that the process $z_l$ will be white since he assumes time independent densities. However, I'm still thinking about the second part of my comment above. – user144410 Feb 10 '17 at 14:40
• Xi'an has more than one MCMC-related book on which he's an author. Please give a full reference, or at the very least a complete, correctly typed title and year (/edition if it's not 1st ed) and a page number – Glen_b Feb 11 '17 at 2:06
• The point is that this sum inside the exponential is of the order of $-t\mathbb{E}_g[\log g(Z)/f(Z)]$ when $t$ is large enough, thus bound to get to $-\infty$ asymptotically, which makes the exponential transform go to zero. – Xi'an Feb 11 '17 at 22:02

Yeah $X_t$ is Markov usuaully, but he's going a step further and asking you to assume these $X_t \overset{iid}{\sim} g$, AND that your model is such that $f$ doesn't need to depend on $t$. Also, he's assuming there is no resampling going on at certain time points. That's what he means when he says "...take the special occurrence where $g_t$ and $f_t$ are independent of $t$..."
It might help to rewrite it like this. Ignore the particle index, and get rid of the $t$s where you don't need them: $$\exp\left\{ t \frac{\sum_{l=1}^t \log[f(z_l)/g(z_l)]}{t} \right\} = \exp\left\{ t \frac{\sum_{l=1}^t r(z_l)}{t} \right\},$$ where $r = \log f - \log g$. So the LLN really does apply.
• Of course, I see this now. That's right. But what happens when we try to take the limit ? We have to do this inside the exponential and only to a part of the argument, namely $\frac{\sum r(z_l)}{t}$. So what happens to that $t$ outside the limit? and what can you imply now on $w_l$ on the left hand side of the proportionality relation? – user144410 Feb 11 '17 at 12:21