Cumulative Distribution Function

For a continuous random variable \(X\), given the cumulative distribution function as shown in the graph below, find \(P(X \le 3.5)\).

Fig. 2 - graph of a cumulative distribution function

Solution:

Don't be fooled because the graph is labelled as

\[ \int f_X(x)\, \mathrm{d}x.\]

Remember that for a probability density function \(f_X(x)\), the integral written is the same thing as the cumulative density function \(F(x)\).

It can help to find the equation of the cumulative distribution function since it is not exactly clear what \(F(3.5)\) is from the picture. Given the graph, you can see that the points \((1,0)\) and \((11,1)\) are the two endpoints of the diagonal line. The equation of that line is then

\[y= \frac{1}{10}x-\frac{1}{10},\]

so you can write down the formula for the cumulative distribution function as

\[ F(x) = \begin{cases} 0 & x < 1 \\ \dfrac{1}{10}x-\dfrac{1}{10} & 1 \le x \le 11 \\ 1 & x > 11 \end{cases}.\]

Now

\[ F(3.5) = \frac{1}{10}(3.5)-\frac{1}{10} = 0.25.\]

In other words, \(P(X \le 3.5) = 0.25\).

Define

\[g(x) = \begin{cases} 0 & x \le 0 \\ \ln x + 1 & x>0\end{cases}.\]

(a) Could \(g(x)\) be a probability density function for a continuous random variable? Explain why or why not.

(b) Could \(g(x)\) be a cumulative distribution function for a continuous random variable? Explain why or why not.

Solution:

(a) For something to be a probability density function, it must always be at least zero. However

\[\begin{align} g\left(\frac{1}{4}\right) &= \ln\left(\frac{1}{4}\right) + 1 \\ &= \ln 1 - \ln 4 + 1 \\ &\approx -0.39 \\ &< 0,\end{align} \]

so it can't be a probability density function.

(b) For something to be a cumulative distribution function, it can't take on any values which are larger than \(1\). However

\[\begin{align} g\left(3\right) &= \ln\left(3\right) + 1 \\ &\approx 2.1 \\ &>1,\end{align} \]

so \(g(x)\) can't be a cumulative distribution function either.

Just because something is written in a piecewise fashion doesn't mean it has anything to do with probability.

Suppose \(X\) is a continuous random variable, and the probability density function is

\[f(x) = \begin{cases} k\sin x & 0 \le x \le \pi \\ 0 &\text{otherwise} \end{cases}.\]

(a) Find the value of \(k\) that makes this work.

(b) Find the associated cumulative distribution function.

(c) Find \(P\left(x \le \dfrac{\pi}{4}\right)\).

Solution:

(a) Remember that for something to be a probability density function, the area under the curve must be equal to one. In other words, you need

\[ \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x = 1.\]

Putting in the function, you get

\[ \begin{align} \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x &= \int_0^\pi k \sin x \, \mathrm{d}x \\ &=\left. -k\cos x \phantom{\frac{}{}} \right|_0^\pi \\ &= -k\cos \pi - (-k\cos 0) \\ &= -k(-1) + k(1)\\ &= 2k. \end{align}\]

So for \(f(x)\) to be a probability density function you need that \(2k = 1\), so \(k = \dfrac{1}{2}\).

(b) From the first part of the problem, you know that

\[f(x) = \begin{cases} \dfrac{1}{2}\sin x & 0 \le x \le \pi \\ 0 &\text{otherwise} \end{cases}.\]

From properties of the cumulative distribution function, you also know that \(F(x)=0\) for \(x \le 0\), and \(F(x)=1\) for \(x \ge \pi\). All that remains is the pesky part between \(0\) and \(\pi\). If you integrate,

\[\begin{align} F(x) &= \int \dfrac{1}{2}\sin x \, \mathrm{d}x \\ &= -\frac{1}{2}\cos x + C \end{align}\]

where \(C\) is the constant of integration. Since \(F(0)=0\),

\[ \begin{align} -\frac{1}{2}\cos x + C &= -\frac{1}{2}\cos 0 + C \\ &= -\frac{1}{2} + C, \end{align} \]

and it must be the case that \(C = \dfrac{1}{2}\). So the cumulative distribution function is:

\[ F(x) = \begin{cases} 0 & x \le 0 \\ -\dfrac{1}{2}\cos x + \dfrac{1}{2} & 0 \le x\le \pi \end{cases}.\]

(c) To find \(P\left(x \le \dfrac{\pi}{4}\right)\), just evaluate \(F\left(\dfrac{\pi}{4}\right) \), giving you

\[ \begin{align} P\left(x \le \dfrac{\pi}{4}\right) &= F\left( \dfrac{\pi}{4}\right) \\ &= -\dfrac{1}{2}\cos \left( \dfrac{\pi}{4}\right) + \dfrac{1}{2} \\ &= -\frac{1}{2}\left(\frac{\sqrt{2}}{2}\right) + \frac{1}{2} \\ & \approx 0.145. \end{align}\]