Suppose that a random variable $Y$ has a p.d.f. given by $$f (y) = ky^3e^{-y/2}$$ when $y > 0$, and otherwise 0. Find the value of $k$ that makes $f(y)$ a density function.
I found that $k=1.$
Does $Y$ have a $\chi^2$ distribution? If so, how many degrees of freedom?
$\nu=8$
What are the mean and standard deviation of $Y$?
Mean is $8$ and standard deviation is $4$.
Could anyone check the answers for me?
Best Answer
You can evaluate the integral over the p.d.f. via integration by parts. (You don't have to completely repeat this procedure when calculating the mean and variance; writing I(n) for the integral over ky^n*e(-y/2), it's easiest to derive a recurrence relation for I(n) in terms of I(n-1).)
Anyway, I disagree with your value of k - but agree with your mean and standard deviation! I'm surprised the error didn't carry over?