Alex's copy of Construct a probability distribution function, then find CDF and expectation

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Alex Van den Hof

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Alex Van den Hof 8 years ago

Created this as a copy of Construct a probability distribution function, then find CDF and expectation.

Name	Status	Author	Last Modified
Construct a probability distribution function, then find CDF and expectation	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Alex's copy of Construct a probability distribution function, then find CDF and expectation	draft	Alex Van den Hof	31/03/2017 11:40
Construct a probability distribution function, then find CDF and expectation	draft	Simon Thomas	07/02/2019 10:43

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Question statement

A random variable $X$ has a probability density function (PDF) given by:

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	0
e1	number	2.5
ux	decimal	dec("8.5")
xl	integer	0
i1	decimal	dec("6")
lo	number	0.08000
j1	decimal	dec("17")
up	number	0.62444
exans	number	0.54444
p	integer	15
u	integer	91
t	integer	60
tol	number	0.01
ans	number	0.54
e2	number	0
cval	number	0.0178
lx	decimal	dec("3")
xu	integer	15

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0

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Part b) Gap-fill
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$f_X(x) = \left \{ \begin{array}{l} \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \\\phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}}\end{array} \right .$	$0$	$ x \leq \var{xl},$

	$cx$	$\var{xl} \lt x \leq \simplify[std]{{xu+xl}/2},$

	$c(\var{xu}-x)$	$\simplify[std]{{xu+xl}/2} \lt x \leq \var{xu},$

	$0$	$x \gt \var{xu}.$

What value of $c$ makes $f_X(x)$ into the pdf of a distribution?

Input your answer here as a fraction and not as a decimal.

$c=\;\;$Unnamed gap

Advice

a)
Note that in order for $f_X(x)$ to be a pdf it must satisfy two important conditions:

1. $f_X(x) \ge 0$ in the range $\var{xl} \le x \le \var{xu}$

2. The area under the curve given by $f_X(x)$ is $1$ and this implies that:
\[\int_{\var{xl}}^{\var{xu}}f_X(x)\;dx = 1\] as the value of the function is $0$ outside this range.

We first check condition 2. and then check that condition 1. is satisfied.

Hence \[\begin{eqnarray*} \int_{-\infty}^{\infty}f_X(x)\;dx=\int_{\var{xl}}^{\var{xu}}f_X(x)\;dx&=&\int_{\var{xl}}^{\simplify[std]{{xu+xl}/2}}cx\;dx+\int_{\simplify[std]{{xu+xl}/2}}^{\var{xu}}c(\var{xu}-x)\;dx\\
&=&c \left[ \frac{x^2}{2} \right]_{\var{xl}}^{\simplify[std]{{xu+xl}/2}} + c\left[\var{xu}x - \frac{x^2}{2}\right]_{\simplify[std]{{xu+xl}/2}}^{\var{xu}}\\&=&\simplify[std]{{(xu+xl)^2}/8}c+\simplify[std]{{(xu-xl)^2}/8}c=\simplify[std]{{xu^2+{xl^2}}/{4}}c \end{eqnarray*} \]

But this has to equal $1$ and so \[c=\simplify[std]{4/{p^2}}\]

Hence with this value of $c$ we see that condition 2. is satisfied i.e.

\[\int_{-\infty}^{\infty}f_X(x)\;dx=1\]

Condition 1. is clearly satisfied as $c \gt 0$.

The Distribution Function

We must have $F_X(x)=0,\;\;\;x \le \var{xl},\;\;\;\textrm{and}\;\;\;F_X(x)=1,\;\;\;x \ge \var{xu}$

Apart from that we find an expression for $F_X(x)$ in each of the ranges $[\var{xl},\simplify[std]{{xl+xu}/2}],\;\;\;[\simplify[std]{{xl+xu}/2},\var{xu}]$

1. $x \in [\var{xl},\simplify[std]{{xl+xu}/2}]$

We have:\[\begin{eqnarray*} f_X(x) &=& \simplify[std]{{4}/{p^2}} x \\ \Rightarrow F_X(x)&=& \int_{-\infty}^xf_X(u)\;du \\ &=& \simplify[std]{{4}/{p^2}}\int_{\var{xl}}^x u\;du\\ &=&\simplify[std]{{4}/{p^2}}\left[\frac{u^2}{2}\right]_{\var{xl}}^x\\&=&\simplify[std]{{2}/{p^2}}x^2 \end{eqnarray*} \]

2. $x \in [\simplify[std]{{xl+xu}/2},\var{xu}]$

We have:\[\begin{eqnarray*} f_X(x) &=& \simplify[std]{{4}/{p^2}}(\var{xu}-x) \\ \Rightarrow F_X(x)&=& \int_{-\infty}^xf_X(u)\;du = \int_{\var{xl}}^{\simplify[std]{{xl+xu}/2}}f_X(u)\;du+\int_{\simplify[std]{{xl+xu}/2}}^x f_X(u)\;du\\ &=& \simplify[std]{{4}/{p^2}}\int_{\var{xl}}^{\simplify[std]{{xl+xu}/2}} u\;du + \simplify[std]{{4}/{p^2}}\int_{\simplify[std]{{xl+xu}/2}}^x(\var{xu}-u)\;du\\&=&\simplify[std]{{4}/{p^2}}\left[ \frac{u^2}{2} \right]_{\var{xl}}^{\simplify[std]{{xu+xl}/2}} + \simplify[std]{{4}/{p^2}}\left[\var{xu}u - \frac{u^2}{2}\right]_{\simplify[std]{{xu+xl}/2}}^x\\&=&1-\simplify[std]{{2}/{p^2}}(\var{xu}-x)^2 \end{eqnarray*} \]

Probability

Using the distribution function we have just found we have that:

$P(\var{lx} \lt X \lt \var{ux}) = F_X(\var{ux})-F_X(\var{lx})$

But $\var{ux}$ is in the range between $\var{(xl+xu)/2}$ and $\var{xu}$ and the distribution function is given by:

\[F_X(x)=1-\simplify[std]{{2}/{p^2}}(\var{xu}-x)^2\]

Hence $F_X(\var{ux})=1-\simplify[std]{{2}/{p^2}}(\var{xu}-\var{ux})^2 = \var{up}$ to 5 decimal places.

Similarly, $\var{lx}$ is in the range between $\var{xl}$ and $\var{(xl+xu)/2}$ and the distribution function is given by:

\[F_X(x)=\simplify[std]{{2}/{p^2}}x^2\]

Hence $F_X(\var{lx})=\simplify[std]{{2}/{p^2}}(\var{lx})^2 = \var{lo}$ to 5 decimal places.

Hence $P(\var{lx} \lt X \lt \var{ux}) = F_X(\var{ux})-F_X(\var{lx})=\var{up}-\var{lo}=\var{up-lo}=\var{ans}$ to 2 decimal places.

Expectation.

d) The expectation is given by $\displaystyle \operatorname{E}[X]=\int_{-\infty}^{\infty}xf_X(x)\;dx$.

As $f_X(x)=0$ for $x \le \var{xl}$ and $x \ge \var{xu}$ we see that:

\[\begin{align}\operatorname{E}[X]= \int_{-\infty}^{\infty}xf_X(x)\;dx &= \int_{\var{xl}}^{\simplify[std]{{(xu+xl)}/2}}xf_X(x)\;dx+ \int_{\simplify[std]{{(xu+xl)}/2}}^{\var{xu}}xf_X(x)\;dx\\&= c\int_{\var{xl}}^{\simplify[std]{{(xu+xl)}/2}}x^2\;dx+c\int_{\simplify[std]{{(xu+xl)}/2}}^{\var{xu}}x({\var{xu}-x)}\;dx\\&=c \left[ \frac{x^3}{3} \right]_{\var{xl}}^{\simplify[std]{{xu+xl}/2}} + c\left[\var{xu}\frac{x^2}{2} - \frac{x^3}{3}\right]_{\simplify[std]{{xu+xl}/2}}^{\var{xu}}\\&=c\times\simplify[std]{{xu}^3/24}+c\times \simplify[std]{{xu}^3/12}\\&=\simplify[std]{{xu}/2}\end{align}\]

on remembering that $\displaystyle c=\simplify[std]{4/{p^2}}$

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Alex's copy of Construct a probability distribution function, then find CDF and expectation

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Alex Van den Hof 8 years ago

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Alex's copy of Construct a probability distribution function, then find CDF and expectation

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

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Checkpoint description

Alex Van den Hof 8 years ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

The Distribution Function

Probability

Expectation.

Generated value: `integer`