Likelihood determination for a step-like pdf

Suppose that random numbers x are generated on the computer using the following procedure:

1. Generate two numbers $$x_1$$, $$x_2$$ from a uniform distribution $$\mathcal{U}$$([0,1])
2. If $$x_1$$ > f, take x = $$x_2$$, otherwise x = m $$x_2$$,

where f and m are two real constants between 0 and 1.

The value of f=0.2 is known, while m is unknown and we want to infer its value from a list of N outputs using the Maximum Likelihood Estimator.

I somehow managed to write the pdf $$p_X(x)$$ for a single observation:

$$p_X(x)=\frac{1}{m(1+f)+f}\mathcal{I}(0,

where $$\mathcal{I}(a is the indicator function.

The likelihood is defined as $$\mathcal{L}=\prod_i p(x_i)$$, and in order to find the MLE i have to solve the known equation $$\frac{d}{dm}\log\mathcal{L}=0$$.

I find some issues in writing the likelihood, because i find myself with

$$\mathcal{L(m|x_1,...,x_N)}=\prod_i p(x_i)=\frac{1}{(m(1+f)+f)^a}+\frac{(1-f)^{N-a}}{(m(1+f)+f)^{N-a}}$$, where $$a$$ is the number of observations $$x_i$$ for which $$x_i. But i added a further unknown parameter to my problem.

I think I am having some issues multiplying the indicator functions, or maybe I have determined the wrong pdf for my problem.

• What does "$a$" represent and how do you manage to convert a product of $N$ quantities into a sum of just two quantities??
– whuber
Sep 3 at 15:45

As $$x_1$$ and $$x_2$$ are independent, the PDF can be written as a mixture of two uniform distributions, one with probability $$(1-f)$$ and support $$(0,1)$$, and the other with probability $$f$$ and support $$(0,m)$$:

$$p(x) = {f \over m}1(x \leq m) + 1 - f$$

As a check, we show this integrates to one:

$$\int_0^1\left({f \over m}1(x \leq m) + 1 - f\right)dx = {f \over m}\int_0^mdx+(1-f)\int_0^1dx = f+(1-f) = 1$$

The multivariate version is:

$$\begin{eqnarray} p(x) &=& \Pi_{i=1}^n\left({f \over m}1(x_i \leq m) + 1 - f\right)\\ &=&\Pi_{i:x_i\leq m}\left({f \over m}+1-f\right)\Pi_{i:x_i>m}(1-f)\\ &\propto&\left({f \over m}+1-f\right)^{n(x \leq m)}(1-f)^{n - n(x \leq m)} \\ &\propto& \left(1+{f \over (1-f)m}\right)^{n(x \leq m)} \end{eqnarray}$$

where $$n(x \leq m)$$ is the number of observations for which $$x \leq m$$. The $$\propto$$ in the last two lines is because we have moved from evaluating the probability based on each $$x_i$$ to evaluating the probability based on $$n(x \leq m)$$; we need to find the constant of integration, which is straightforward:

$$p(x) = {\left(1+{f \over (1-f)m}\right)^{n(x \leq m)} \over \sum_{i=0}^n\left(1+{f \over (1-f)m}\right)^i}$$

Given that $$m$$ appears in the term $$n(x \leq m)$$ in the exponent as well as in the $$(1 + f/((1-f)m))$$ terms in both numerator and denominator and that $$n(x \leq m)$$ is a piecewise horizontal function with $$n$$ jumps, I suspect finding $$\hat{m}$$ by numerical methods is probably the way to go.

• Many thanks for the answer! I thought it would look something like this. But i am pretty sure that there is an analytical solution to the MLE of m. Sep 4 at 7:19
• The problem is the jumps in $n(x \leq m)$ as you increase $m$; calculus won't help you with that. Sep 4 at 14:32