I've known there are only two kinds of the variables that discrete and continuous.
So I met the big problem bothering me.
Q) Let $F(x)$ be c.d.f(cumulative distribution function) for the variable $x$ like the below. Find the constant, $k $
$F(x) =\begin{cases}
0, & \text{$x \lt0$}\\
{1 \over 2} x^2& \text{$0 \leq x \lt 1$}\\
k(4x-x^2), & \text{$1 \leq x \lt 2$} \\
1 & \text{$x \geq 2$}
\end{cases}$
When I saw it first, I regarded as the continuous variable so we can find the $f$, p.d.f.(probability density function) by $F'(x) = f(x)$
$f(x) =\begin{cases}
0, & \text{$x \lt0$}\\
x& \text{$0 \leq x \lt 1$}\\
k(4-2x), & \text{$1 \leq x \lt 2$} \\
\end{cases}$
Therefore, $k = {1\over 2}$ since $\int_0^{2} f dx = 1$
But the answer was $k = {1\over 4}$
What did I wrong? Is the variable $x$ not continuous random variable?(Surely it is not discrete in my guess since $[0,2]$ itself uncountable)
Best Answer
There is no density for this $F$. You have to just apply the definition of a distribution function. You will see that $F$ meets the requirements of a distribution function iff $\frac 1 6 \leq k \leq \frac 1 4$.
A distribution function is a non-decreasing function $F: \mathbb R \to \mathbb R $ such that $F(x) \to 0$ as $x \to -\infty$, $F(x) \to 1$ as $x \to \infty$ and $F(y)\to F(x)$ as $y$ decreases to $x$ for every real number $x$.
There are discrete random variables, there are random variables with densities and there are many other types of random variables. We may have $P(X=x)=0$ for every real number $x$ (so that $X$ is not discrete) but still $X$ many not have a density.