# Likelihood function & MLE without known values of observed data

Question: Let $X_1,\dots,X_n$ be iid exponential rate $\lambda$. Suppose we don't know the observed values of our experiments, but we know that $k$ values were $\le M$ and the remaining $n-k$ were $>M$ for some constant $M$. Find the MLE of $\lambda$.

I need to find the likelihood function $L(\lambda; \vec x)$, which is typically defined as $\prod_{i=1}^n f(x_i;\lambda)$ for iid data. Now, I intuitively believe that the $L(\lambda; \vec x)$ will equal ${n \choose k} [P(X \le M)]^k [P(X > M)]^{n-k}$, but I'm not satisfied with my explanation why.

The likelihood function is defined as $L(\theta) = f(x_1,\dots,x_n;\theta)$, where $\theta$ is allowed to vary. I don't know how to relate this definition in terms of the pdf to the fact that precisely $k$ observed values were $\le M$. Can someone help with a more rigorous explanation?

If you don't know the actual observed values but only the number of observations not exceeding a certain value, your data is actually a realisation $k$ of an integer-valued random variable $K \in \{0, 1, \ldots, n\}$.

Furthermore, since the sample is i.i.d. the whole experiment can be thought of as constituting of independent replications of observing a value for $X_i$ and then recording only whether it is $\leq M$ or $>M$. Introducing a corresponding indicator variable $K_i$ that takes the value 1 in the former case and 0 in the latter case, the random variable corresponding to the data,

$$K = \sum_{i=1}^n K_i,$$

is a sum of independent Bernoulli-distributed indicator variables, and as such binomially distributed with the "success probability" $P(X \leq M)$. That is, $K \sim Bin(n, P(X \leq M))$ and this leads directly to the likelihood function postulated in the question.

• Thank you - a great explanation. So they key was to view the experiment in a new fashion; namely, to look at it as a sequence of independent Bernoulli trials, correct? I was simply trying to manipulate the definition $L(\theta) = f(x_1,\dots,x_n;\theta)$ to get this likelihood, and I couldn't get anywhere. – user365239 Jan 11 '17 at 20:32
• @user365239 Exactly. Using the exponential likelihood doesn't work here as you don't actually observe any exponentially distributed data, just binomial. – J. Virta Jan 11 '17 at 22:00