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say we have a random sample of a random variable, but we are told that no information is available for values less than or equal to some number, perhaps k. That is, we don't have the full sample. We only get the values that are greater than k. So, how do we account for this in the likelihood function? Do we multiply by (1 - F(k)) or something like that? Any help is greatly appreciated!

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Supposing that $k$ is known, if you start with a sample density $f(x_i\mid\theta)$, with support on the real line, for each $\theta$, you may define, for your truncated sample, a new sample density $$ g(x_i\mid\theta) = \frac{f(x_i\mid\theta)}{Z(\theta)} I_{(k,\infty)}(x_i) \, , $$ where $Z(\theta)=\int_k^\infty f(t\mid\theta)\,dt$. Hence, the likelihood is $$ L_{x_1,\dots,x_n}(\theta) = \frac{\prod_{i=1}^n f(x_i\mid\theta)}{Z^n(\theta)} \, , $$ where I have discarded the product of indicators, which does no depend on the parameter $\theta$.

If $k$ is unknown, and must be treated as an additional parameter, then you have a different expression for the likelihood of $(\theta,k)$. Can you write it down?

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  • $\begingroup$ So divide by $(1 - F(k))^n$. $\endgroup$
    – GeoffDS
    Apr 28 '13 at 21:20
  • $\begingroup$ Yeah. The problem is that generally you don't have an analytic expression for this denominator (which is a function of $\theta$). $\endgroup$
    – Zen
    Apr 28 '13 at 21:24
  • $\begingroup$ Just to be sure I understand even more, since I am no expert, you are using a truncated density. The truncated density starts at our value $k$ but the actual density would only integrate to $1 - F(k)$ if we integrated it from $k$ to $\infty$. Thus, we divide the truncated density by this value so it integrates to 1. Similarly, if we truncated it above at $k$, we would use the original density, but only defined from $0$ to $k$, and divide that by $F(k)$. Is that all right? $\endgroup$
    – GeoffDS
    Apr 28 '13 at 22:05
  • $\begingroup$ You mean "but only defined from $-\infty$ to $k$"? Everything else sounds right. $\endgroup$
    – Zen
    Apr 28 '13 at 22:10
  • $\begingroup$ Sure, a detail I didn't mention in this problem is that the problem I was working on involved an exponential distribution. And, I'm studying for an actuarial exam where most (maybe all) of the distributions involved are only defined starting at 0. $\endgroup$
    – GeoffDS
    Apr 29 '13 at 0:05

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