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$p(x)=\frac{1}{9}(x-1)$, $x=2,3,4,5$. Why can this function not be a p.m.f?

$f(x)=\frac{2}{3}(x-3)$, $2\leq x\leq5$ Why can this function not be a p.d.f?

1 Answers 1

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Hint

Is the total mass or probability $1$ in each case; and are the functions non-negative?

In case $1$, is it true that $$\sum_{i=2}^5p(x)\overset{?}{=}1$$

In case $2$, is it true that $$\int_2^5f(x)\,\mathrm{d}x\overset{?}=1$$

$p(x)$ is not a probability mass function and $f(x)$ is not a probability density function.

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    There is no mention of mass or probability2012-05-02
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    I have edited to add what I mean, does it look ok to you?2012-05-03
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    Actually, in case 2, the integral is 1. However, the integrand takes negative values...2012-05-03
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    In case 1, yes it does = 1 since: p(2)=1/9 p(3)=2/9 p(4)=3/9 p(5)=4/9. When taking the sum of these values it indeed does equate to 1.2012-05-03
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    @kate No, it sums to 10/9.2012-05-03
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    Far too late at night for this.. sorry for the silly arithmetic mistake! :) x2012-05-03