Home page for accesible maths 2.4 Continuous random variables 2.4.1 The probability density function 2.5 The Quantile function

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An illuminating idea

For some very small interval width $\delta$ ,

	$\displaystyle\operatorname{\mathsf{P}}\left({x<X\leq x+\delta}\right)$	$\displaystyle=F_{X}(x+\delta)-F_{X}(x)$
		$\displaystyle=\int_{x}^{x+\delta}f_{X}(s)\,\mathrm{d}s$
		$\displaystyle\approx f_{X}(x)\delta.$

Thus $f_{X}(x)\delta$ can be thought of as (approximately) the probability that $X$ is between $x$ and $x+\delta$ .

On rearranging we obtain

\displaystyle f_{X}(x)\approx[F(x+\delta)-F(x)]/\delta\approx\,\mathrm{d}{F_{X% }}/\,\mathrm{d}{x},

with the approximations becoming exact in the limit as $\delta\rightarrow 0$ . This illustrates the equivalence of (2.3) and our definition of $f_{X}(x)$ .

Example 2.4.1.

A random variable $X$ has cumulative distribution function

\displaystyle F_{X}(x)=\left\{\begin{array}[]{ll}0&\quad x\leq 0\\ x&\quad 0<x\leq 1\\ 1&\quad x>1\end{array}\right.

Find the pdf of $X$ .

Solution.

\displaystyle f_{X}(x)=\frac{\,\mathrm{d}}{\,\mathrm{d}x}F_{X}(x)=\left\{% \begin{array}[]{ll}0&\quad x\leq 0\\ 1&\quad 0<x\leq 1\\ 0&\quad x>1\end{array}\right.

Example 2.4.2.

A triangular pdf: a random variable $X$ has pdf

\displaystyle f_{X}(x)=\left\{\begin{array}[]{ll}1+x&\quad-1<x\leq 0\\ 1-x&\quad 0<x\leq 1\\ 0&\quad\text{otherwise}\end{array}\right.

Obtain the cdf $F_{X}(x)$ .

Solution. Important: We split the range of $x$ , $(-\infty,\infty)$ into sensible intervals.

•

For $x\in(-\infty,-1]$ ,

$\displaystyle F(x)=\int_{-\infty}^{x}f_{X}(s)\,\mathrm{d}s={\color[rgb]{% 0.76,0.01,0}\int_{-\infty}^{x}0\,\mathrm{d}s=0.}$
•

For $x\in{\color[rgb]{0.76,0.01,0}(-1,0],}$

$\displaystyle F(x)$ $\displaystyle=\int_{-\infty}^{x}f_{X}(s)\,\mathrm{d}s$

$\displaystyle={\color[rgb]{0.76,0.01,0}F(-1)+\int_{-1}^{x}1+s\,\mathrm{d}s}$

$\displaystyle={\color[rgb]{0.76,0.01,0}0+\left[s+s^{2}/2\right]_{-1}^{x}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}x+x^{2}/2+1/2=(1+x)^{2}/2.}$
•

For $x\in{\color[rgb]{0.76,0.01,0}(0,1],}$

$\displaystyle F(x)$ $\displaystyle=\int_{-\infty}^{x}f_{X}(s)\,\mathrm{d}s$

$\displaystyle={\color[rgb]{0.76,0.01,0}F(0)+\int_{0}^{x}1-s\,\mathrm{d}s}$

$\displaystyle={\color[rgb]{0.76,0.01,0}1/2+\left[s-s^{2}/2\right]_{0}^{x}}$

$\displaystyle={\color[rgb]{0.76,0.01,0}1/2+x-x^{2}/2=1-(1-x)^{2}/2.}$
•

For $x\in{\color[rgb]{0.76,0.01,0}(1,\infty),}$

$\displaystyle F(x)=\int_{-\infty}^{x}f_{X}(s)\,\mathrm{d}s={\color[rgb]{% 0.76,0.01,0}F(1)+\int_{1}^{x}0\,\mathrm{d}s=1.}$

Hence

\displaystyle F_{X}(x)=\left\{\begin{array}[]{ll}{\color[rgb]{0.76,0.01,0}0}&% \quad x\leq-1\\ {\color[rgb]{0.76,0.01,0}(1+x)^{2}/2}&\quad-1<x\leq 0\\ {\color[rgb]{0.76,0.01,0}1-(1-x)^{2}/2}&\quad 0<x\leq 1\\ {\color[rgb]{0.76,0.01,0}1}&\quad x>1\end{array}\right.

Example 2.4.3.

Find $\operatorname{\mathsf{P}}\left({-0.5<X<2}\right)$ , where $X$ is the random variable from the previous example.

Solution. $F_{X}(2)-F_{X}(-0.5)=1-(1-0.5)^{2}/2=7/8$ .

	$\displaystyle F(x)$	$\displaystyle=\int_{-\infty}^{x}f_{X}(s)\,\mathrm{d}s$
		$\displaystyle={\color[rgb]{0.76,0.01,0}F(-1)+\int_{-1}^{x}1+s\,\mathrm{d}s}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}0+\left[s+s^{2}/2\right]_{-1}^{x}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}x+x^{2}/2+1/2=(1+x)^{2}/2.}$

	$\displaystyle F(x)$	$\displaystyle=\int_{-\infty}^{x}f_{X}(s)\,\mathrm{d}s$
		$\displaystyle={\color[rgb]{0.76,0.01,0}F(0)+\int_{0}^{x}1-s\,\mathrm{d}s}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1/2+\left[s-s^{2}/2\right]_{0}^{x}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1/2+x-x^{2}/2=1-(1-x)^{2}/2.}$