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The question is:

The continuous probability distribution function (PDF) depends on one parameter $a$ related to the slope of the line and can be defined as: $$ \operatorname{pdf}(x | a)=\left\{\begin{array}{ll}{1+a x,} & {\text { for }-\frac{1}{2} \leq x \leq \frac{1}{2}} \\ {0,} & {\text { otherwise }}\end{array}\right. $$ a) What is the range of values for $a$ to ensure that the above definition is a valid probability distribution function?

My answer is:

$0 \leq p d f(x | a) \leq 1$

So: $\quad 1+a x \geqslant 0$ and $1+a x \leq 1$

$1^{\circ}$ for $-\frac{1}{2} \leq x \leq \frac{1}{2} \quad$ and $\quad a>0$ $\quad $ $\quad\left\{\begin{array}{l}{1+a x \geqslant 0} \\ {1+a x \leq 1}\end{array}\right\} \cdot \frac{1}{a} \rightarrow \left\{\begin{array}{l}{x \geqslant-\frac{1}{a}} \\ {x \leq 0}\end{array} \Rightarrow-\frac{1}{2}=-\frac{1}{a} \Rightarrow a=2\right.$

$2^{\circ}$ for $-\frac{1}{2} \leq x \leq \frac{1}{2} \quad$ and $\quad a<0$ $\quad $ $\quad\left\{\begin{array}{l}{1+a x \geqslant 0} \\ {1+a x \leq 1}\end{array}\right\} \cdot \frac{1}{a} \rightarrow \left\{\begin{array}{l}{x \leq-\frac{1}{a}} \\ {x \leq 0}\end{array} \Rightarrow\frac{1}{2}=-\frac{1}{a} \Rightarrow a=-2\right.$

$\Rightarrow a \in[-2,2]$

Is this answer correct? I feel like this isn't the correct approach but it's been so long since I last did inequalities.

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    $\begingroup$ If you are calling it a "pdf" you may also want $\int_{-0.5}^{0.5} f(x|a) dx = 1$, which will determine admissible values of $a$. $\endgroup$
    – snar
    Oct 23 '19 at 16:07
  • $\begingroup$ Oh! Now I understand! So the pdf itself can take values greater than 1, but the area should be 1. Thank you super much! $\endgroup$
    – saremisona
    Oct 23 '19 at 16:12
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pdf must take nonnegative values but it can take values that are bigger than $1$.

Hence we need $1-\frac12 a \ge 0$ and $1+\frac12 a \ge 0$. We just need to check the boundary value since it is a straight line segment.

which is equivalent to $2 \ge a$ and $a\ge -2$.

That is $$-2 \le a \le 2$$

Remark: It is easy to check that integration to $1$ condition holds trivially.

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  • $\begingroup$ How come it can take values bigger than 1? Isn't the maximum probability 1? I am missing something big here. It's been too long since I took a probability course. $\endgroup$
    – saremisona
    Oct 23 '19 at 16:10
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    $\begingroup$ For example, consider the uniform distribution between $0$ to $\frac12$. The values that the pdf value take is not the probability, the integration gives you the probability. $\endgroup$ Oct 23 '19 at 16:12
  • $\begingroup$ Finally I understand! I knew there are something basic I was missing. Thank you super much! $\endgroup$
    – saremisona
    Oct 23 '19 at 16:12

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