Suppose that a random variable $Y$ has a probability density function given by
f(y) = \begin{cases} ky^3e^{-\frac{y}{2}}, & y>0 \\ 0, & \text{elsewhere} \end{cases}
(a) Find the value of $k$ that makes $f(y($ a density function.
(b) Does $Y$ have a chi-square distribution? If so, how many degrees of freedom?
(c) What are the mean and standard deviation of $Y$?
Solution
(a) $\int ^{\infty}_{-\infty} f(y)$ $dy=1$
$=\int^{\infty}_0 y^3e^{-\frac{y}{2}}$ $dy=\frac{1}{k}$
I obtained
$e^{-\frac{y}{2}}[-2y^3-12y^2-48y-6]^{\infty}_0=\frac{1}{k}$
What happens when $y$ tends to $\infty$?
Foe (b), I do not know.
For (c), I will be able to manage if my part (a) is explained.
For all $n$, $\lim\limits_{y\to\infty}e^{-y/2}y^n = 0$
The pdf of a Chi-squared distribution of $\kappa$ degrees of freedom is : $$f_Y(y)~=~\dfrac{y^{\kappa/2-1}e^{-y/2}}{2^{\kappa/2}\Gamma(\kappa/2)}$$