Question

Integrating ky(1-y), where k is an unknown constant and the integral = to 1 between infinity and negative infinity

Answer 1

k should be 6, but I am unsure about how exactly that comes to be

Answer 2

what is between infinity and negative infinite ??

Answer 3

This is a density function for a continuous stochastic variable, f(y) = ky(1-y), 0 <= y <= 1

I need to find k, for this function to be a probability density function

Answer 4

sorry ... I have no experience with stochastic calculus. i thought it was elementary calculus question.

Answer 5

lol

Answer 6

The stochastic part I can handle, it is the calculus that is confusing me. It should integrate to 1 in the end, and I just need the procedure to integrate that bad boy

Answer 7

If it's just improper integration then maybe i could help
could you type your equation on here
http://www.codecogs.com/latex/eqneditor.php and post/paste it ... or use equation tool to type equation?

Answer 8

http://latex.codecogs.com/gif.latex? \int%20ky(1-y)dy%20=%201

Answer 9

is this the equation?
\[ \int_0^\infty ky(1-y) dy\]

Answer 10

\[\int\limits_{ - \infty}^{\infty}ky(1-y) dy = 1\] Excuse my ignorance, I'm used to pencil and paper

Answer 11

this integral does not converge I don't think it has Cauchy Principle Value either.

Answer 12

I know that k = 6, for the function to integrate to 1. But thank you anyway. I will go scribble on a wall somewhere to figure it out.

Answer 13

sure ... good luck.

Answer 14

Um, looking at the supposed answer, I think the integral should be
\[\int_0^1 ky(1-y) dy\]
Since \(0 \leq y \leq 1\)

Then it would be
\[
k * \left(\frac{y}{2} - \frac{y^3}{3}\right)\bigg|^1_0=\frac{k}{6}
\]

Then k would be 6 for the integral to evaluate to 1..