Determining constant in a CDF

Question

I have a question that I literally have no idea how to begin, I was hoping someone could help me:

1. It says $X_1,X_2,\ldots,X_n$ is a sample from a distribution

2. It says that the Cumulative Distribution Function (CDF) is

$$F(x,\theta) = 1 - C/((x-\theta)+3)^4,\quad x > 0, \theta > 0$$

The instructions now say

determine the constant C (no computation required) and determine the pdf f(x).

Can anyone point me in a direction of a resource? or give me a hint as to how to get started? I have no clue what this distribution function is at all!

Answer 1

If $F(x,\theta) = 1-\frac{C}{((x-\theta)+3)^4}$ is a cumulative distribution function, then we know that it must satisfy certain properties, namely:

1. $F(x,\theta) \to 0$ as $x\to -\infty$;
2. $F(x,\theta) \to 1$ as $x\to \infty$;
3. $\frac{dF}{dx} \ge 0\ \forall x$.

The probability density function is simply $\frac{dF}{dx}$, by definition. To find the constant $C$, we need only look at the first few properties. Hint: for any arbitrary choice of $\theta$, what does $\frac{1}{((x-\theta)+3)^4}$ look like?

Answer 2

Let $F(x,\theta)$ be defined for $x>0$ and $\theta < 3$ by $$F(x,\theta) = 1 - {C\over(x-\theta+3)^4}\,.$$ If $F$ is a CDF in $x\in (0, \infty)$, then what must be the value of $C$?

Well, as $x\rightarrow 0$ from the right we have $$F(x,\theta)\rightarrow 1 - {C\over (3-\theta)^4}\ .$$ For $F$ to be a CDF, this limit must be zero, so that $C = (3-\theta)^4$.

That makes $$F(x,\theta) = 1 - {(3-\theta)^4\over(x + (3-\theta))^4}\,.$$

Note that $F\rightarrow 1$ as $x\rightarrow \infty$, as required for a CDF. It can also be verified that ${\partial f\over\partial x} \geq 0$.