From the book Introduction to Mathematical Statistics by Hogg, McKean and Craig (# 6.1.12):

Let $X_1,X_2,\cdots,X_n$ be a random sample from a distribution with one of two pdfs.

If $\theta=1$, then $f(x;\theta=1)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\,-\infty<x<\infty$.

If $\theta=2$, then $f(x;\theta=2)=\frac{1}{\pi(1+x^2)},\,-\infty<x<\infty$. Find the mle of $\theta$.

My attempt: derive the first $f$ with respect to $x$ and set it to zero. That gives me $x_{1}=0$. Replacing in the first $f$, we get $\sqrt{\frac{1}{2 \pi}}$.
Working similarly with the second $f$ we get the value of $\frac{1}{\pi}$.
The former is greater, the the final answer is $\theta = 1$.

Is that right? If not, what would be the right procedure?

  • 1
    $\begingroup$ You seem to be thinking right but doing it all wrong. You should be using $X_1, X_2, \ldots, X_n$. Remember, these samples came from one of the above two pdfs. If you had to decide which pdf it came from, how will you go about doing it? $\endgroup$ Jun 29, 2018 at 3:51
  • 1
    $\begingroup$ You need to be calculating likelihoods in there somewhere. $\endgroup$
    – Glen_b
    Jun 29, 2018 at 4:15
  • $\begingroup$ You might want to add the self-study tag and read it's wiki. $\endgroup$ Jun 29, 2018 at 5:59
  • $\begingroup$ @StubbornAtom Got it, I just added the self-study tag. I also noticed that you typed out the question, which I originally posted as a screenshot. Is that what I should always do? $\endgroup$
    – evaristegd
    Jun 30, 2018 at 14:04
  • $\begingroup$ stats.stackexchange.com/q/145014/119261 $\endgroup$ Feb 17, 2020 at 14:39

1 Answer 1


In this problem your unknown parameter $\theta$ only has two possible values, so you have a discrete optimisation where you just have to compare the likelihood at those two parameter values. (If you are taking derivatives of something in a discrete optimisation then you are going down the wrong track.) For an observed data vector $\mathbf{x}$ you have:

$$L_\mathbf{x}(\theta) = \begin{cases} (2 \pi)^{-n/2} \exp(-\tfrac{1}{2} \sum x_i^2) & & \text{for } \theta = 1, \\[6pt] \pi^{-n} / \prod (1+x_i^2) & & \text{for } \theta = 2. \\ \end{cases}$$

Since there are only two possible parameter values, you can find the maximising parameter value by looking at the sign of the difference in likelihood at these values. You have:

$$\begin{equation} \begin{aligned} \Delta(\mathbf{x}) \equiv \text{sgn}(L_\mathbf{x}(1)-L_\mathbf{x}(2)) &= \text{sgn}\Bigg( (2 \pi)^{-n/2} \exp(-\tfrac{1}{2} \sum x_i^2) - \frac{1}{\pi^{n} \prod (1+x_i^2)} \Bigg) \\[6pt] &= \text{sgn}\Bigg( \exp(-\tfrac{1}{2} \sum x_i^2)\prod (1+x_i^2) - \Bigg( \frac{2}{\pi} \Bigg)^{n/2} \Bigg). \\[6pt] \end{aligned} \end{equation}$$

The maximum-likelihood-estimator (MLE) is:

$$\hat{\theta} = \begin{cases} 2 & & \text{if } \Delta(\mathbf{x}) = -1, \\[6pt] \{ 1,2 \} & & \text{if } \Delta(\mathbf{x}) = 0, \\[6pt] 1 & & \text{if } \Delta(\mathbf{x}) = 1. \\[6pt] \end{cases}$$

(In the case where $\Delta(\mathbf{x}) = 0$ the MLE is non-unique since the likelihood is the same at both parameter values.)

  • $\begingroup$ Thank you. When you go form the first line to the second one, after multiplying both terms by $[\Pi (1+x_i)^2]^2$ I get $sgn((exp(-\frac{1}{2} \sum x_i^2))(\Pi (1+x_i^2))-(\frac{2}{\pi})^{n/2})$. Did I get something wrong? $\endgroup$
    – evaristegd
    Jun 30, 2018 at 14:19
  • $\begingroup$ Yes, that is correct - edited. $\endgroup$
    – Ben
    Jul 1, 2018 at 2:01

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