Let $f_0$ and $f_1$ be two continuous probability density functions with means $\mu_0,\mu_1$ and variances $\sigma_0^2,\sigma_1^2$ on $\mathbb{R}$. Furthermore, let $l(y)=f_1(y)/f_0(y)$ be the likelihood ratio and $0<c_l<1$ and $c_u>1$ be two positive real numbers.
Each time, we draw $n$ i.i.d. samples $(y_1,y_2...,y_n)$ from the distribution $f_0$ and calculate the vector $v=[\ln l(y_1),\ln l(y_2),...,\ln l(y_n)]$. If any element of $v$ is larger than $\ln c_u$ then they are replaced by $\ln c_u$ and similarly if any element of $v$ is smaller than $\ln c_l$ those are changed by $\ln c_l$, giving another vector $v^*$.
Example: For $v=[-3.1, 0.6, -7.5, 0 ,4.8, 2.4, .1]$, $\ln c_l=-2$ and $\ln c_u=3$ we get $v^*=[-2, 0.6, -2, 0 ,3, 2.4, 0.1]$. The values $−3.1$ and $−7.5$ are clipped to $-2$ and $4.8$ is clipped to $3$.
Consider the sum: $$s_n=\sum_{i=1}^n v^*(i)$$ To which distribution does the empirical distribution of $s_n$ converge as $n\rightarrow\infty$? (probably in terms of $c_l,c_u$ and the means and variances of $f_0$ and $f_1$)
Added:(14.04.2014)
To get the density of $s_n$ we need to first know the density of each $v^*_i$. Since the there is clipping the density of $v^*_i$ will be defined on the interval $[\ln c_l, \ln c_u]$. Out of this interval the density of $v^*_i$ will be zero. at exactly $\ln c_l$, the density of $v^*_i$ will have a point mass which is $\int_{\{y:\ln l<\ln l_l\}}f_0(y)\mbox{d}y$ and similarly at $\ln c_u$, there is a point mass of $\int_{\{y:\ln l>\ln l_u\}}f_0(y)\mbox{d}y$. Between $\ln c_l$ and $\ln c_u$, the likelihood doesn't change and I think the density of the likelihood will stay the same at that range (should be checked).
Does anyone have any idea?
Thank you very much!!!
(Not an answer, too long for comment. ) I doubt that you can express that in terms of the moments (mean, variance) of $f_0$ and $f_1$.
If we forget for one moment about the "clipping", we have the random variable
$$z=\log \frac{f_1(y)}{f_0(y)}$$ where $y$ follows the density $f_0(y)$. Then, we know its mean: it's the Kullback Leibler distance (or divergence, or relative entropy).
$$E(z)= - \int f_0(y) \log \frac{f_0(y)}{f_1(y)} dy=-D( f_0 || f_1)$$
Then we can expect that $s_n/n \to -D( f_0 || f_1)$ (but we need to check the convergence conditions, and which convergence). Anyway, this is not expressable in terms of means and variances of $f_0$ and $f_1$.
If we can bound the second moment of $z$ (but againg I doubt that it have simpler expression that the mere integration), we can apply then the CLT to $s_n/\sqrt{n}$. Our problem is complicated by the clipping, of course.