Construct PDF and find CDF,

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Newcastle University Mathematics and Statistics

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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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
valk	decimal	dec("0.6667")
xl	integer	1
p	integer	1
pval	decimal	dec("0.42")
xu	integer	2

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Name

Data type

Value

xxxxxxxxxx
 
0

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Description

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Generated value: `integer`

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pval
valk

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Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
Part c) Gap-fill
1. Gap Mathematical expression
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Gap-fill

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$f_X(x) = \left \{ \begin{array}{l} \phantom{{.}} \\ \phantom{{.}} \\ \phantom{{.}} \end{array} \right .$	$kx$	$\var{xl} \leq x \leq \var{xu},$

	$0,$	$\textrm{otherwise.}$

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

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

$k=\;\;$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.

Note that \[\int kx\;dx = k\frac{x^2}{2}\] on forgetting the constant of integration.

Hence \[\begin{eqnarray*} \int_{\var{xl}}^{\var{xu}}kx\;dx &=&\frac{k}{2}(\var{xu}^2-\var{xl}^2)\\ &=&\frac{k}{2}\times \var{xu^2-xl^2} \end{eqnarray*} \]

But we must have this last value equal to $1$ hence:
\[ \frac{k}{2}\times \var{xu^2-xl^2}=1 \Rightarrow k = \simplify[std]{2/{xu^2-xl^2}}\]

Hence the pdf is:
\[f_X(x) = \simplify[std]{2/{xu^2-xl^2}x}\;\;\;\;\;\var{xl} \le x \le \var{xu}\]

We have to check condition 1. that the function $f_X(x)$ is positive for $\var{xl} \le x \le \var{xu} $ – but this is clear from
the definition of $f_X(x)$ and the value of $k$.

The Distribution Function

If $F_X(x)$ is the distribution function of the distribution given by $f_X(x)$ then:

$F_X(x) = 0\;\;\;x \lt \var{xl},\;\;\;\;F_X(x)=1\;\;\;x \ge \var{xu}$

and for $\var{xl} \le x \le \var{xu}$:

\[\begin{eqnarray*} F_X(x)&=&\int_{-\infty}^x f_X(x)\;dx=\simplify[std]{2/{xu^2-xl^2}}\int_{\var{xl}}^x x\;dx\\ &=&\simplify[std]{2/{xu^2-xl^2}}\times\frac{\left(x^2-\var{xl}^2\right)}{2}\\ &=&\frac{x^2-\var{xl^2}}{\var{xu^2-xl^2}} \end{eqnarray*} \]

We have
\[\begin{eqnarray*} P\left(X \lt \simplify[std]{{xl+xu}/2}\right)&=&F_X\left(\simplify[std]{{(xl+xu)}/2}\right)\\ &=& \frac{1}{\var{xu^2-xl^2}}\left(\simplify[std]{({(xl+xu)}/{2})^2-{xl}^2}\right)\\ &=&\simplify{{3*xl+xu-4*a}/{4*(xu+xl-2*a)}} \end{eqnarray*} \]

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

The Distribution Function

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Construct PDF and find CDF,

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

The Distribution Function

Generated value: `integer`