For which parameters $a,b \in \mathbb{R}$ function $F(t)$ are distribution of random variable?

Question

For which parameters $a,b \in \mathbb{R}$ function $$F(t)= \begin{cases} b - \frac{a}{(t+1)^{2}}, & \text{for } t \geq - \frac{1}{2} \\ 0 & \text{for } t < -\frac{1}{2} \end{cases}$$ are distribution of random variable?

Idea with issue:

From properties of distribution, we have:

1. $\lim_{n\to \infty}F(t)=1 $
2. $\lim_{n\to -\infty}F(t)=0 $
3. F(t) is monotonically increasing function.

From 1. We get $b=1$ and from 2. have b=0

We compute $a$ for $b =1 $ with properties 3. $$1- \frac{a}{(-\frac{1}{2}+1)^2}>0$$ so $$ a < \frac{1}{4} $$

For $b = 0$ $$ a < 0 $$

But I don't think my solution is right, because we have tuple of parameters. Do you have any proposition?

Answer 1

You are making a mistake when tending $t\to-\infty$ from the first equation. From $(1)$ we can say $b=1$. For $t\to-\infty$, the function is zero for small enough $t$. Now note that ${1\over (1+t)^2}$ is a decreasing function of $t$; therefore for $a$ being non-negative we have a monotonically increasing function and we must have $$0\le b-4a\le 1\implies 0\le a\le {1\over 4}$$If the continuity of CDF matters to us, then $$b-4a=0\implies a={1\over 4}$$

Answer 2

The mistake you are making is that $\lim_{t \to -\infty} F(t)$ is already $0$ for any choice of $a, b$ since the function is identically $0$ on $(-\infty, - 1/2)$.

You need $b = 1$ to ensure $\lim_{t \to \infty} F(t) = 1$. Notice that $a \geq 0$, otherwise $F(t)$ gets bigger than $1$. Assuming $a > 0$ the function given is already non-decreasing on $[-1/2, \infty)$. Finally, you have to check if $F(-1/2) \geq 0$, which is similar to what you had found $a \le 1/4$.

So, the function $F$ represents the distribution of a random variable whenever $b=1$ and $a \in [0, 1/4]$.