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A random variable $X$ is represented by the following CDF: $F(x)=(1+\frac{1}{x^2})^{-\beta}$ , $x\in(0, \infty), \beta>0$

Question: How do you get the MLE of $P(X>1)$ for the distribution?

I thought of two ways:

1) I tried to transform the CDF by subtracting $-F(1)$ from $F(x)$, so the CDF would represent only the values for $x\in(1, \infty)$. But this yields the same MLE as for $P(X)$. Can this be possible respectively is this approach wrong?

2) Can i just add $1$ to every value of $x$ in the CDF or will this transform the CDF incorrectly?

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    $\begingroup$ Could you clarify what you mean by "MLE $\hat{\beta}$ of $P(X<1)$"? You are asking for the MLE of some number between 0 and 1 and this doesn't make sense. $\endgroup$
    – TrynnaDoStat
    Mar 30, 2015 at 11:37
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    $\begingroup$ Is this a self-study exercise? Either way, the notation is muddled here. It's conventional that $\hat\beta$ is taken to denote the MLE of $\beta$. An estimator of $P(X < 1)$ follows from plugging in $x = 1$ but would not itself be $\hat\beta$. $\endgroup$
    – Nick Cox
    Mar 30, 2015 at 11:46
  • $\begingroup$ I edited the post to make it more clear. It is not a self-study exercise. Original exercise was "give the MLE of P(X>1) for the distribution", what i thought was bad phrasing. $\endgroup$
    – Fusscreme
    Mar 30, 2015 at 11:53
  • $\begingroup$ The edits don't fix the problem. The phrasing you were given is better. $\endgroup$
    – Nick Cox
    Mar 30, 2015 at 16:37
  • $\begingroup$ The problem really is simpler than your formulation. The CDF is the same throughout the problem: there is no need to think of any constraints or any transformations. You just need to read off one point on the CDF. The fact that the problem is one of maximum likelihood estimation is separate. $\endgroup$
    – Nick Cox
    Mar 31, 2015 at 10:13

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You should use the invariance of ML estimators.

What you are looking for is $\widehat{P(X>1)}$, where $P(X>1) = 1 - (1+\frac{1}{1^2})^{-\beta}$. That is, you are estimating an injective function of beta. So, it can be computed as $1 - (1+\frac{1}{1^2})^{-\hat\beta}$, again, because of the invariance of the MLE.

So your excercise is just calculating $\hat\beta$, the MLE of $\beta$ and plugging it into the expression as commented above.

If my calculations are right, this last is $\hat\beta = \frac{n}{\sum_{i=1}^n{ln(1+\frac{1}{x_i^2})}}$, but correct me if I am wrong.

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  • $\begingroup$ How did you obtain that formula for the MLE? For the case $n=1$ I get $\hat\beta = 1/\log(1+1/x^2)$, which is quite different. $\endgroup$
    – whuber ♦
    Mar 30, 2015 at 20:03
  • $\begingroup$ I get the same MLE as whuber. And I know about the invariance of ML estimators. However, I do not understand how the invariance of ML estimators can be used when you do not change the parameters ($\beta$) but alter the constraint ($P(X>1)$) of the CDF. $\endgroup$
    – Fusscreme
    Mar 30, 2015 at 20:39
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    $\begingroup$ Fusscreme, this approach is correct (+1; although the MLE is still wrong, I believe). Because $P(X\gt 1)=1-2^{-\beta}$, the MLE of $P(X\gt 1)$ must be $1-2^{-\hat\beta}$ when $\hat\beta$ is the MLE of $\beta$. See stats.stackexchange.com/questions/80584, for instance. $\endgroup$
    – whuber ♦
    Mar 30, 2015 at 23:04
  • $\begingroup$ @whuber I checked it (the one in my edited answer) and I keep getting to the same result (and in n = 1 it is the same as yours). Should I remove it in case I am wrong? Since it is not the center of the argument. $\endgroup$
    – Cristián Antuña
    Mar 31, 2015 at 1:13
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    $\begingroup$ I apologize; your new MLE is correct. (I originally had not recognized a simplification of the log likelihood that leads to this nice formula.) $\endgroup$
    – whuber ♦
    Mar 31, 2015 at 15:30

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