Probability function question

Hi there,

Given a probability function \(\displaystyle P(Y=y) = \left\{\begin{array}{ccc}\frac{1}{9}|y| \\ 0 \ \mathrm{other}. \end{array}\)

where y belongs to the the interval ]-3,3[

I would like show that P(|y| <= 1/9) = 1/9

Any hints or idears to how I do that ?

Do I make a seperate interval \(\displaystyle P(y \in ]-3,-1/9[ \cup ]1/9, 3[) = ?\) ?

Sincerely Yours and God bless You all

Sophie

Draw a picture and compute the area under the graph.

Okay then if p is continuous in ]-3,3[. Then for each y in ]-3,3[.


\(\displaystyle P(|y| \leq \frac{1}{9}) = \int \limit_{-3} ^{y} P\)

Then by the fundamental theorem of Calculus P is differentiable in ]-3,3[ and \(\displaystyle P(|y| \leq \frac{1}{9})^{\prime} = \frac{1}{9}\)

Does this sound right?

Best Regards

Sophie Toft