Suppose $X_1, X_2, \ldots, X_n$ are $IID$ normal RVs with mean $\mu$ and variance $1$. However, we observe only $Y_i$'s where $Y_i = \max (0, X_i)$. I would like to know how to write likelihood function for $\mu$ given $Y_i$'s. Since, $Y_i$ has both discrete and continuous components, I tried to write it in the following way

$$L(\mu \mid Y_1,Y_2,\ldots, Y_n) = \frac{1}{(2\pi)^\frac{m}{2}}\exp\left(-\frac{1}{2}\sum_1^m (Y_i-\mu)^2 \right) \left(\Phi(-\mu)\right)^{n-m}$$

where $Y_1, Y_2, \ldots, Y_m > 0$ and $Y_{m+1}=Y_{m+2}=\cdots=Y_n=0$ and $\Phi$ is CDF of standard normal.

I am not sure if this is correct way of writing it. I would be thankful if anyone can direct me to any references on how to write likelihood functions when the distribution of data has both discrete and continuous components.

  • 1
    $\begingroup$ Normally if one says "data has both discrete and continuous components" I would take that to mean it's something like $(X_i,Y_i),\ i=1,\ldots,n$ where $X_i$ has a discrete distribution and $Y_i$ has a continuous distribution. That's not what you have here. The distribution of $Y_i$ is actually a mixture, i.e. a weighted average, of a discrete distribution and a continuous distribution. That discrete distribution concentrates probability $1$ at $0$. The weight assigned to it depends on the value of $\mu$. $\qquad$ $\endgroup$ – Michael Hardy Feb 5 '16 at 18:36
  • 2
    $\begingroup$ I am inclined to agree with your likelihood function, but I'm to rushed to write a thoughtful justification right now. $\qquad$ $\endgroup$ – Michael Hardy Feb 5 '16 at 18:56
  • $\begingroup$ Please give justification when you have some time! Thanks $\endgroup$ – chandu1729 Feb 6 '16 at 23:19

Let's assign a measure $m$ to Borel subsets of the half-open interval $[0,\infty)$ by specifying that the measure of every open interval is its length and $m(\{0\})=1$, and measures of all other Borel sets are accordingly determined. Let $f$ by a probability density with respect to the measure $m$, so that \begin{align} & \int_{[0,\infty)} f(x)\,dm(x) = \int_{(0,\infty)} f(x)\,dm(x) + \int_{\{0\}} f(x)\,dm(x) \\[10pt] = {} & \int_{(0,\infty)} f(x)\,dm(x) + f(0)m(\{0\}) = \int_{(0,\infty)} f(x)\,dm(x) + f(0). \end{align} For a random variable $X$ having this distribution, we have $$ \Pr(X=0) = \int_{\{0\}} f(x)\,dm(x) = f(0)m(\{0\}) = f(0). $$ Let $$ f_\mu(x) = \begin{cases} \displaystyle \varphi(x-\mu) & \text{if } x\ne 0, \\[10pt] \Phi(-\mu) & \text{if }x=0, \end{cases} $$ (where $\Phi$ and $\varphi=\Phi'$ are the standard normal c.d.f. and p.d.f. respectively). This is the density for the observations you describe, and you can define the likelihood function $\mu\mapsto L(\mu)$ accordingly.

Let us see what would happen if we had used a different measure with respect to which our desired probability distribution has a density. What if we had said $m(\{0\})=1/2$ and left the rest of the definition as above? Then we would have $$ L(\mu) = \frac{1}{(2\pi)^{m/2}}\exp\left(-\frac{1}{2}\sum_1^m (Y_i-\mu)^2 \right) \left(2\Phi(-\mu)\right)^{n-m} $$ with the additional factor of $2^{n-m}$. But it is still proportional to $$ \mu \mapsto \exp\left(-\frac{1}{2}\sum_1^m (Y_i-\mu)^2 \right) \left(\Phi(-\mu)\right)^{n-m}. $$ With likelihood functions, the proportionality class is all that matters, and a certain amount of seeming arbitrariness in the choice of the initial measure does not change that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.