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Evo8
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Homework Statement
$$P_x(x)=A(1- \frac{|x|}{2}) \ \ \ |x|≤2$$
$$P_x(x)=0 \ \ |x|>2$$
Find A
Homework Equations
The Attempt at a Solution
This one shouldn't be too bad but I wanted to verify that I am on the right track.
I basically have ##P_x(x)=A(1- \frac{x}{2})## when x≤2. 0 otherwise.
So I write the integral ##\int_{-\infty}^{2} A(1-\frac{|x|}{2})dx##
I end up with ##A (x(\frac{-x^2}{4})dx## evaluated from 2 to ##-\infty## which will equate to ##\infty##
The section in my text that talks about probability distribution function mentions this
$$\int_{-\infty}^{\infty} p_x(x)dx=1$$
So I could do ##A\int_{-\infty}^{\infty} 1-\frac{|x|}{2} dx=1##? Solving this would yield A=##\infty## which is not what I am looking for I don't think. Its something to do with the absolute value of x i think...
The form that the question writes the equation in always screws me up.
Thanks for any help