This is a qualitative question. In the literature of particle filters/ sequential monte carlo, particle degeneracy is unavoidable. However, frequented cases where it may happen is when the likelihood is very peaked. The result of this


Degenracy is usually explained as the all the weights becoming extremely small and it becomes very sensitive due to the peaked likelihood. Therefore, when we normalise we get all the weights accumulated by 1 particle or a few particles from the entire particle set. Then it is state that this idea of degeneracy corresponds to the $var(w^i_t)$ being extremely high.I do not get the logical deduction that degeneracy ->var(w^i_t) becoming high.

Do they mean that the value of $w^i_t$ is extremely sensitive, meaning a slight nudge to particle $x^i_t$ in a particular direction(closer to the likelihood peak) will result in the $w^i_t$ having high weights all of a sudden. Or could it mean that if I repeated an experiment for another time that $w^i_t$ is likely to have a different value?

Can someone give a mathematical formulation of this?


1 Answer 1


Redefine $$ \frac{p(y_t|X^i_t)p(X^i_t|x^i_{t-1})}{q(X^i_t|x^i_{t-1})} = h(X_t^i). $$ My hope is that this will emphasize that the weight adjustment random variable is a transformation of your new sample. Also, the data $y_t$ is assumed constant throughout this.

Peakedness versus Variance

The term $\text{Var}(h(X_t^i)|w_{t-1}^i)$ measures how spread out this weight adjustment random variable is. This is related to the range of possible values $h(X_t^i)$ can take. If $p(y_t|X_{t}^i)$ is ``peaked," thought of as a function of $X_t^i$, it can take on a wider range of values. This means $h(X_t^i)$ is peaked as well, and it can take on more values. Of course the variance depends on the probability of taking on big values, but ignore that for a minute. Also, the variance of $h(X_t^i)$ has to do with $p(X_t^i|x_{t-1}^i)$ and $q(X_t^i|x_{t-1}^i)$, but these are densities, so they usually don't influence the peakedness of $h(X_t^i)$ that much.

Peakedness Versus Weight Histogram

If $p(y_t|X_{t}^i)$ is ``peaked," thought of as a function of $X_t^i$, then there are very few values of $X_t^i$ that can yield extreme values of $p(y_t|X_t^i)$, which in turn yields extreme values of $h(X_t^i)$. This can be seen if you plot the weight, $h(X_t^i)$ against $X_t^i$. If you draw $X_t^i \sim q(x_t^i|x_{t-1}^i)$, then most of these draws will be associated with a low weight. Think of the graph you drew for the above situation. Occasionally, though, you will draw a sample that maps to the peaked region of $h(X_t^i)$. This is what gives you the left-skewed histogram.

  • $\begingroup$ I truly do not understand why if the likelihood is peaked that it can take a wider range of values? Shouldnt it be that if the likelihood is peaked that it can take a larger range of values. Unless you mean that there is a larger range of values with smaller density(hand-wavy). And is $y^i$ intended to be the incremental weight or the un-normalised weight at time $t$? $\endgroup$ Mar 30, 2017 at 0:04
  • $\begingroup$ @tintinthong these are unnormalized weights. Can you edit your first two sentences? Peaked as a function of $x_t$. $\endgroup$
    – Taylor
    Mar 30, 2017 at 0:32
  • $\begingroup$ fair enough. You mean that likelihood is a function of $X^i_t$, a random sample. So, the likelihood is random in the sense that it is conditioned on a random variable $X^i_t$. I am still a little confused. What is your definition of "peaked"? If $y^i$ are the unnormalised weights then where is $w^i_{t-1}$ in the equation your re-defined $\endgroup$ Mar 30, 2017 at 15:19
  • $\begingroup$ @tintinthong see edits. I had bad notation in the sense that I was using the letter $y$ in multiple ways. Peaked means that it's pointy, so $h$ is peaked if it's flat most places except for a few values where it's very high. If you're sampling inputs to this function, you're mostly likely going to get a place where $h$ is low. But one or two samples will get a very high weight adjustment. I dropped $w_{t-1}^i$ because I'm taking the conditional variance, so it's a constant. This is not always done; sometimes people think of the current weight as a function of all past weight adjustments. $\endgroup$
    – Taylor
    Mar 30, 2017 at 15:57
  • 1
    $\begingroup$ @tintinthong if $h(X_t^i)$ is peaked (thinking of it as a function of $X_t^i$) then there is a big range on it. Then think of the histogram of $h(X_t^i)$...it will be spread out more--so higher variance. $\endgroup$
    – Taylor
    Mar 30, 2017 at 17:19

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