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I am trying to find the maximum likelihood estimate. I did but, I stack at the last part, how can draw $a$?

Log likelihood function is as follows

\begin{align} \ell & =\ln(L(a))=\ln \left( \prod_{i=1}^n (1/\sqrt{2\pi}\sigma)(1/\sqrt{y_i-a}))\exp\left(-\frac{y_i-a}{2\sigma^2}\right)\right) \\[10pt] & =n \ln (1/\sqrt{2\pi}\sigma)+\sum^n_{i=1} \ln(y_i-a)^{-1}+\sum_{i=n} \left(-\frac{y_i-a}{2\sigma^2}\right) \end{align}

Let's take the derivative with respect to $a,$ I get the following

\begin{align} & =\sum^n_{i=1}(y_i-a)^{-1} +\sum^n_{i=1} 1/2\sigma^2 \\[10pt] & =\sum^n_{i=1}(y_i-a)^{-1} + n/2\sigma^2=0 \end{align}

How can I derive $a$ alone from the following equation?

$$\frac{n}{2\sigma^2}+ \sum_{i=1}^n (y_i-a)^{-1} =0$$ where $y_i \not= a$

I would like to write $a=\ldots$

This is the last part of my main question. Since I'm stuck at this point, I only wrote this part. My attempt is as follows:

$$n\prod_{i=1}^n(y_i-a)+2\sigma^2[[(y_2-a)\cdots(y_n-a)]+[(y_1-a)(y_3-a)\cdots(y_n-a)]+\cdots+[(y_1-a)(y_2-a)\cdots(y_{n-1}-a)]]=0$$

But I also cannot derive $a$ from this equation.

Because I am not studying math, I don't know summation calculation techniques.

Thank you.

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  • $\begingroup$ You wrote "where $y_i>a.$ Why is that? There seems to be something you're not telling us. $\endgroup$ – Michael Hardy Nov 17 '17 at 19:54
  • $\begingroup$ It appears that you are considering the probability distribution of $(Y-a)^2$ where $Y\sim N(a,\sigma^2).$ But you haven't said that. $\endgroup$ – Michael Hardy Nov 17 '17 at 20:10
  • $\begingroup$ It seems to be something of a knee-jerk reflex to seek the value that makes the derivative $0$ and then to think that's all there is to the problme of finding a global maximum. But even in first-semester calculus you are told that there is such a thing as a maximum at and endpoint. And that is just what you're dealing with here. $\endgroup$ – Michael Hardy Nov 17 '17 at 20:18
  • $\begingroup$ You wrote "How can I derive $b$ alone[...]?" But you haven't told us what $b$ is. $\endgroup$ – Michael Hardy Nov 17 '17 at 20:20
  • $\begingroup$ I did not understand exactly. Can you show what you said a bit in mathematical expression? @MichaelHardy $\endgroup$ – user489188 Nov 17 '17 at 20:21
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If $Y= a + X^2$ and $X\sim N(0,\sigma^2),$ then we have the density $$ f_Y(y) = \frac 1 {\sqrt{2\pi (y-a)}} \cdot \frac 1 \sigma \exp\left( \frac{-(y-a)}{2\sigma^2} \right) \text{ for } y \ge a. $$ You didn't tell us that $Y=a+X^2$ and $X\sim N(0,\sigma^2)$ was the scenario you were considering, but the form of the density you wrote conjoined with your later comment that $y\ge a$ seems to imply it.

You have $$ \ell = n \ln \frac 1 {\sqrt{2\pi}\sigma} + \sum^n_{i=1} \ln(y_i-a)^{-1} + \sum_{i=n} \left(-\frac{y_i-a}{2\sigma^2}\right)$$

and you want the value of $a$ that maximizes that, subject to the constraint $a\le y_i$ for each $i.$ That means $a \le \min\{y_1,\ldots,y_n\}.$

You have $$ \frac{\partial\ell}{\partial a} = \sum^n_{i=1} \frac 1 {y_i-a} +\sum^n_{i=1} \frac 1 {2\sigma^2}. $$ Therefore $\dfrac{\partial \ell}{\partial a}>0$ for all values of $a$ in the domain, which is the interval $(-\infty,\min\{y_1,\ldots,y_n\}].$ That means the maximum value of $\ell$ occurs when $a$ is at the upper endpoint of the interval $(-\infty,\min\{y_1,\ldots,y_n\}],$ i.e. we have $$ \widehat a = \min\{y_1,\ldots,y_n\}. $$

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  • $\begingroup$ Yes you are right y=a+x^2 $\endgroup$ – user489188 Nov 17 '17 at 21:39
  • $\begingroup$ okay I could not think such a result. I am very happy that you help me in this point. Thanks a lot:) $\endgroup$ – user489188 Nov 17 '17 at 21:40
  • $\begingroup$ Dear Michael Hardy I have also such a question. Please can you look at if you mind math.stackexchange.com/questions/2525332/… $\endgroup$ – user489188 Nov 17 '17 at 21:49

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