Deriving MLE of $\theta$ in $\text{Exp}(\theta,\theta)$ distribution

Suppose $ X_1,X_2,\ldots,X_n$ are i.i.d variables having a two-parameter exponential distribution with common location and scale parameter $ \theta$ :

$ $ f_{\theta}(x)=\frac{1}{\theta}\exp\left[-\left(\frac{x-\theta}{\theta}\right)\right]\mathbf1_{x>\theta}\quad,\,\theta>0$ $

I am wondering if it is possible to derive a maximum likelihood estimator (MLE) of $ \theta$ .

The likelihood function given the sample $ x_1,\ldots,x_n$ is

$ $ L(\theta)=\frac{1}{\theta^n}e^{-n(\bar x-\theta)/\theta}\mathbf1_{x_{(1)}>\theta}\quad,\,\theta>0$ $

, where $ \bar x=\frac{1}{n}\sum\limits_{i=1}^n x_i$ and $ x_{(1)}=\min\limits_{1\le i\le n} x_i$ .

Since $ L(\theta)$ is not differentiable at $ \theta=x_{(1)}$ , I cannot apply the second-derivative test here.

Even if I could say that $ L(\theta)$ is increasing and/or decreasing in $ \theta$ under the constraint $ \theta<x_{(1)}$ , I am not sure what choice of $ \theta$ maximises $ L(\theta)$ . Differentiation is not valid as I understand.

If MLE is unique, then it is bound to be a function of the sufficient statistic $ (\overline X,X_{(1)})$ . However I don’t see how to derive it in this particular model. Any suggestion would be great.

Combinatorics: Distribution of red, blue and green balls into distinct bins

Assume that there are three different kinds of balls, R red balls, B blue balls, and G green balls. How many ways are there to distribute the balls into N different bins so that no bin is empty and a bin can have at most one ball of the same color? Here,

1) R,B,G<=N (no color should have more than N balls as we have the condition that at most one balls of the same color in a bin and all balls needs to be placed)

and

2) R+B+G>=N (This is due to the fact that no bin is empty).

Proof of Poisson process distribution

I’m confused about part of a proof of this theorem.

Theorem: If {$ N_t; t \geq 0$ } is a Poisson process, then for any $ t \geq 0$ , $ $ P(N_t = k) = \frac{e^{\lambda t}(\lambda t)^k}{k!}$ $ where $ k = 0, 1, …$ and some constant $ \lambda \geq 0$ .

Proof

Let $ G(t) = E(\alpha ^{N_t})$ . Writing $ N_{t+s} = N_t + (N_{t+s} – N_t)$ , using the independence of $ N_{t+s} – N_t$ [and prior results] we get $ $ G(t+s) = E(\alpha ^{N_{t+s}})=E(\alpha ^{N_t}\alpha ^{N_{t+s}-N_t})=E(\alpha ^{N_t})E(\alpha ^{N_{t+s}-N_t})=G(t)(G(s)$ $ Since $ G(t)=\sum \alpha ^n P(N_t = n) \geq P(N_t = 0)=e^{-\lambda t}, G$ does not vanish for any $ t$ , and $ G(t+s) = G(t)G(s)$ can be satisfied only if

$ $ G(t) = e^{tg(\alpha)}, t \geq 0$ $

Note that $ g(\alpha)$ is the derivative of $ G$ at $ t=0$

I’m confused about the previous two lines. Why does $ G(t)$ need to have the form above? And how is it that $ g(\alpha)$ is the derivative of $ G$ at $ 0$ ? It seems like circular referencing, $ g$ is the derivative, yet it appears in the expression of $ G$ . Strange.

what is the difference between the pseudorandom proportional rule and probability distribution in the ACO system

When we applied the pseudorandom proportional rule and probability distribution we find the same destination (next city), so what is the importance the both rules ? if cannot understand the main difference between them ? if you have any exampe to how use the rules o more details. https://imgur.com/SiKIOgT

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How to determine the distribution of Ubuntu

I want to launch an EC2 instance on AWS. I want to install Ubunto Xenial distribution. How do I know which one of these is Xenial?

I see a lot of options like:

  • Ubuntu Server 18.04 LTS (HVM), SSD Volume Type – ami-0b76c3b150c6b1423
  • Ubuntu Server 14.04 LTS (HVM), SSD Volume Type – ami-001dae151248753a2
  • Deep Learning AMI (Ubuntu) Version 23.0 – ami-0726ab58f406b644f
  • Deep Learning Base AMI (Ubuntu) Version 18.0 – ami-00c2ec90e50ed2f33
  • Ubuntu Server 16.04 LTS (HVM) with SQL Server 2017 Standard – ami-f13ff693
  • Ubuntu Server 14.04 LTS (HVM), SSD Volume Type – ami-0d21bd76bdbb39f53

Update

The reason I want Xenial is that Galera cluster is already built for xenial:

http://releases.galeracluster.com/mysql-wsrep-5.6.39-25.22/ubuntu/dists/xenial/

Nesting Email Distribution Groups into job code Groups

My company wants to create job code groups. A single group for a single job code. For example, if Bob is a courier, the only groups he will have is the “Domain User” and Courier. Couriers are part of 3 distribution groups, 1 universal and 2 global. They are also in 4 global security groups and 1 mail-enabled global security group.

Of course, no reason is given, just they want it that way.

The problem I am facing is the end user is not receiving the distribution email.

Any suggestions?

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