Cumulative distribution function to probability density

Question

I am given the cumulative distribution function $F(x) = 1-e^{\frac{-x^2}{2 \alpha}}$ for $x>0$. $X$ describes how long a component works before it fails. $\alpha$ is a parameter describing the quality of the components.

I need to find the probability density of $X$. I tried taking the derivative of $F(x)$, because if it were the other way around (find cumulative distribution $F(x)$ given density function $f(x)$), I would integrate from $0$ to $\infty$. I get a nasty expression however, and the answer is supposed to be $\sqrt{\alpha}$

Have I misunderstood the relationship between cumulative distribution and probability density?

EDIT: "The cumulative distribution function is shown in the figure below, for the case $\alpha = 1$. From the figure we see that it's very probable that the component will fail after 5 years"

Answer 1

You are correct that the relationship between the probability density function $f$ and cumulative distribution function $F$ is that $f = F'$.

Are you sure the answer is supposed to be $\sqrt{\alpha}$? This would mean the density does not depend on $x$ (which doesn't make sense).

EDIT: If you are looking for the value which maximizes the probability density function, this is $x = \sqrt{\alpha}$.

To find this, first compute the density function, $$ f(x) = F'(x) = \frac{xe^{-\frac{x^2}{2\alpha}}}{\alpha} $$

Now find the maximum of $f(x)$ for $x\geq 0$ by solving $f'(x) = 0$.

Answer 2

If your CDF is given by

$$ F(x;\alpha) =\begin{align}\begin{cases} 1 - e^{\frac{-x^{2}}{2\alpha}} & x > 0 \\ \\ 0 & x \leq 0 \end{cases} \end{align} \tag{1}$$

then we know that

$$ \lim_{x \to \infty} F(x;\alpha) = 1 \tag{2}$$

if we plug that in then

$$ \lim_{x \to \infty} F(x;\alpha) = 1 \tag{3}$$

given that $\alpha >0$. If you take $\frac{d}{dx}F(x)$ you'll get the pdf.

$$ f(x;\alpha) =\begin{align}\begin{cases} \frac{xe^{\frac{-x^{2}}{2\alpha}} }{a} & x > 0 \\ \\ 0 & x \leq 0 \end{cases} \end{align} \tag{4}$$

The Rayleigh Distribution is

$$ f(x;\sigma) = \frac{x}{\sigma^{2}}e^{\frac{-x^{2}}{2 \sigma^{2} }} \tag{5} x \geq 0$$

with CDF

$$ F(x;\sigma) =\begin{align}\begin{cases} 1 - e^{\frac{-x^{2}}{2\sigma^{2}}} & x > 0 \\ \\ 0 & x \leq 0 \end{cases} \end{align} \tag{6}$$

It's a rayleigh distribution.