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Unit 02 /
Problem 3:


E cdf of a continuous rv
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Easy


Let $F(x)=3x-2x^2$ for $0\leq x\leq1$, $F(x)=0$ for $x\leq0$, and $F(x)=1$ for $x\geq1$. Is $F$ a valid cumulative distribution function?

Solution:

For $F$ to be a CDF, it must fulfill 3 properties.

  1. $F$ must be monotonically increasing.
  2. $\lim_{x \to - \infty} F(X) = 0$
  3. $\lim_{x \to \infty} F(X) = 1$

Since $F(x) = 1$ for any $x \geq 1$, $\lim_{x \to \infty} F(X) = 1$. Since $F(x) = 0$ for any $x \leq 0$, $\lim_{x \to - \infty} F(X) = 0$.

To see if $F$ is monotonically increasing, we suggest that you try graphing it to get some intuition. Since $F$ is continuous, we can take its derivative and check if it's always nonnegative.

$$ \frac{d}{dx} F(x) = \frac{d}{dx} 3x-2x^2 = 3 - 4x$$

We can notice quickly that there is an inflection point at $x=\frac{3}{4}$, and a quick spot-evaluation (without even using any further calculus identities) shows that for $\frac{3}{4} \leq x \leq 1$ this function is decreasing. And so, it does not meet the definition of a valid cdf.