# Find the original pdf given conditions in order to find Maximum Likelihood Estimator

Random Sample Distribution from $X_1,X_2,...,X_n$.

Given the p.d.f of $f(x;\theta)$

Find the Maximum Likelihood Estimator $\theta$;

When $\theta = 0$, $f(x;\theta) = 1$ where $0<x<1$. When $\theta=1$, $f(x;\theta) = \frac{1}{{2}\sqrt{x}}$ where $0 < x < 1$.

I am having trouble finding the equation of this statement. Originally, the pdf is not given. But, probably wrong, I THINK the equation is

$$f(x;\theta) = \frac{1}{2^\theta}x^{-\theta/2}$$

However, if this is the equation, the maximum likelihood I calculated turns out to be $0$. I find this really strange.

Could someone give some advice of how I could find the original pdf given those conditions so that I can solve the MLE?

• I've answered the question, explaining the likelihood function, which I hope helps. By the way I think there's a slight error in your post for $\theta=1$: it should be $\frac1{2\sqrt{x}}$. – TooTone Feb 14 '14 at 14:01
• I've amended my answer after your edit, but I'm still not sure it's helping you. I really don't understand how you could be asked to find the maximum likelihood estimator without knowing the pdf (or being told the name of the distribution, which is equivalent). – TooTone Feb 14 '14 at 16:30

## 1 Answer

Could someone give some advice of how I could find the original pdf given those conditions so that I can solve the MLE?

You can't. In general, a probability density has to be non-negative and sum to 1. So given any non-negative function $g(x)$, you can convert it to a probability density by dividing it by its sum over the range of $x$. Put

$$c = \int_{-\infty}^\infty g(x)\;dx$$

Then

$$f(x) = \frac{1}{c}g(x)$$

is a probability density because it's non zero and its sum is $\frac{1}{c} \int_{-\infty}^\infty g(x)\;dx = \frac{1}{c}c = 1$.

$g(x)$ is known as the core. Your two conditions when $\theta=0$, $f(x;\theta)=1$, and when $\theta=1$, $f(x;\theta)=1/(2\sqrt{x})$, where $0<x<1$, aren't sufficient to determine $g(x)$.

In any case,

\begin{align} \int_0^1 f(x;\theta) \;dx &= \frac{1}{2^\theta} \int_0^1 x^{-\theta/2} \;dx\\ &= \frac{1}{2^\theta} \times \frac{2}{2-\theta}\left[ x^{1-\theta/2}\right]_0^1,\;\;\;\theta\neq2\\ &= \frac{1}{2^{\theta-1}(2-\theta)} \end{align}

This is equal to $1$ for $\theta=0$ and $\theta=1$ but isn't equal to 1 in general, so $f(x;\theta)$ isn't, in general, a probability density.

• $$f(x;\theta) = \frac{1}{2^\theta}x^{-\theta/2}\;\;\;\text{where }0 < x < 1$$ This is not actually the pdf given. I just guessed that it would be this equation. So does that mean in this case it would be L(θ) = 1 x 1/(2√x)? – Max Feb 14 '14 at 14:59
• Well I still don't have any idea how to solve this problem, but I did learn something else that helped me with another question so its still worth it. I am still stuck with this question though – Max Feb 15 '14 at 7:43