Likelihood function and MLE Please let me know how to find the likelihood function and MLE for the function $$f(x;θ) = (θ+1)(x^θ)$$
I have tried using the general formula for likelihood function $L(θ)$ however not sure how to proceed further.
Please assist.
 A: Let $X=(X_1,\dots,X_n)$ with the $X_i$ i.i.d with pdf $f(x;\theta)$.
Then
$$L(\theta;X)=\prod_{i=1}^nf(X_i;\theta)=(\theta+1)^n(X_1\dots X_n)^\theta$$
It is very often better to consider the logarithm of the likelihood function:
$$LL(\theta;X)=n\log(\theta+1)+\theta\sum_{i=1}^nX_i$$
Then you can differentiate this with respect to $\theta$ to find the MLE.
A: You need to state the support of this function.
Clearly, you can't mean that your density function
has this form over the entire real line.
I suppose you intend to say this is the density
function for $x \in (0,1)$ and it's 0 elsewhere.
If so, this distribution is a member of the beta
family of distributions. Look at the Wikipedia
article on 'beta distribution' for the answer to your question (and more,
which you can read of ignore as you prefer). See especially,
Sect. 3 "Parameter estimation: maximum likelihood".
Beta is a
two-parameter family. In your case, figure out what the
first parameter is in terms of your $\theta$; the second
parameter is $\beta = 1.$
A: The likelihood for a single observed $x$ is $L(\theta) = (\theta+1) x^\theta$.  I.e., the density is the expression on the right as a function of $x$ with $\theta$ fixed, and the likelihood is the same expression as a function of $\theta$ with $x$ fixed.
Often exercises involve an i.i.d. sample.  In that case, the likelihood is
$$
L(\theta) = \underbrace{\Big((\theta+1)x_1^\theta\Big) \cdots \cdots \Big( (\theta+1) x_n^\theta \Big)}_{\text{Just multiply.}} = \underbrace{(\theta+1)^n (x_1\cdots x_n)^\theta}_{ \begin{smallmatrix} \text{Then do algebraic} \\ \text{simplifications.} \end{smallmatrix} }.
$$
(Presumably the support would be $x\in[0,1]$. That should be mentioned when stating the problem.)
A: The likelihood function you are looking for is
$$L(x_1,\dots,x_n|\theta)=(\theta+1)^n\left(\prod_{i=1}^nx_i\right)^{\theta}$$
Taking the natural logarithm on the expression above gives
$$\ln (L(x_1,\dots,x_n|\theta))=\ln\left((\theta+1)^n\left(\prod_{i=1}^nx_i\right)^{\theta}\right)=n\ln(\theta+1)+\theta\left(\sum_{i=1}^n\ln x_i\right)$$
Now, differentiating with respect to $\theta$ to find the maximum gives
$$\frac{d}{d\theta}\ln (L(x_1,\dots,x_n|\theta))=\frac{n}{\theta+1}+\sum_{i=1}^n\ln x_i=0$$
So we just need to solve the equation for $\theta$. That gives us 
$$\hat{\theta}=\frac{-n}{\sum_{i=1}^n\ln x_1}-1$$
which is the $MLE$ of the function.
