Skip to main content

[Community Question] Statistics: Can a summation be transferred into the denominator?

One of our user asked:

I include a bit of an introduction, even though my main question is more mathematical. I was tasked with finding the Maximum Likelihood Estimate for $\theta$ in $$\mathrm P(X>x) = \left(\frac ax \right)^\theta $$ where $X$ is a variable, and $x$ represents a value that variable can take on. The Probability Density Function is $\newcommand\diff[2]{\frac{\mathrm d#1}{\mathrm d#2}}\diff Fx=\frac{-\theta a^\theta}{x^{\theta + 1}}$, where $F = \mathrm P(X>x)$. I maximise the loglikelihood function $l = \ln(-\theta) + \theta \ln a - (\theta + 1)\ln x\ $ to get $\hat\theta(x_i) = \frac 1{\ln x_i - \ln a}$, where the $\hat.$ indicates that $\hat\theta$ is an estimate of $\theta$, based on the data sample. Now, the answer is supposed to be $$\hat\theta = \frac 1{\overline {\ln x} - \ln a}$$ where $\overline {\phantom{x}}$ indicates the average: $\overline{\ln x} = \frac 1n \sum_i \ln x_i$. I am stumped as to how to get this answer directly from $\hat\theta(x_i)$.

Does $$\frac 1n \sum_i \frac 1{\ln x_i - \ln a} = \frac 1{\overline {\ln x} - \ln a}\qquad ?$$

I think $\frac 1n \sum_i \widehat{\frac 1{\theta(x_i)}} = \frac 1n\sum_i (\ln x_i - \ln a) =\overline{\ln x} - \ln a = \widehat {\frac 1\theta} \implies \hat\theta = \frac 1{\overline{\ln x} - \ln a}$, but is this the only way to show the above?


Labels:

Comments

Post a Comment

Popular posts from this blog

[Community Question] Calculus: Asymptotic behavior of $\sum\limits_{n=0}^{\infty}x^{b^n}$

One of our user asked: This follow my previous post here , where Song has proven that $\forall b>1,\lim\limits_{x\to 1^{-}}\frac{1}{\ln(1-x)}\sum\limits_{n=0}^{\infty}x^{b^n}=-\frac{1}{\ln(b)}$ , that is to say : $$\forall b>1,\sum\limits_{n=0}^{\infty}x^{b^n}=-\log_b(1-x)+o_{x\to1^-}\left(\log_b(1-x)\right)$$ (The $o_{x\to1^-}\left(\log_b(1-x)\right)$ representing a function that is asymptotically smaller than $\log_b(1-x)$ when $x\to1^{-}$ , that is to say whose quotient by $\log_b(1-x)$ converges to $0$ as $x\to1^{-}$ , see small o notation ) So we have here a first asymptotical approximation of $\sum\limits_{n=0}^{\infty}x^{b^n}$ . I now want to take it one step further and refine the asymptotical behaviour, by proving a stronger result which I conjecture to be true (backed by numerical simulations) : $$\sum\limits_{n=0}^{\infty}x^{b^n}=-\log_b(1-x)+O_{x\to1^-}\left(1\right)$$ (The $O_{x\to1^-}\left(1\right)$ representing a function that is asymptotically bounded wh...
Post a Comment
Read more

[Community Question] Linear-algebra: Explicit example of an additive map which is not R-linear

One of our user asked: Is there an explicit example of an additive map $\mathbb{R}^n \rightarrow \mathbb{R}^m$ which is not linear? (I have mostly thought about the question when $m = n = 1$ , and I don't think the general case is any easier.) I know that something like $f: \mathbb{C} \rightarrow \mathbb{C}$ which sends $f: z \mapsto \text{Real}(z)$ would be additive but not $\mathbb{C}$ -linear. I also know that since $\mathbb{R}$ is a $\mathbb{Q}$ -vector space, I can find some example where $1 \mapsto 1$ and $\sqrt{2} \mapsto 0$ . Is there an explicit example? By explicit, I mean, given an element in the domain, there would be some procedure to decide where it maps. Thank you!
Post a Comment
Read more