# Thread: Continuous random variables and their PDFs

1. ## Continuous random variables and their PDFs

The question goes:

Suppose the continuous random variable X has a probability density function (PDF)

$\displaystyle f_{X}(x)=\frac{c}{x^6}, x>1$

i) Show that for X to be a valid PDF, $\displaystyle c$ must be equal to $\displaystyle 5$.
ii) Calculate $\displaystyle E(X^6e^{-2X})$.

For part i) I keep getting $\displaystyle -5$, not $\displaystyle 5$ for $\displaystyle c$, and I just can't fathom out part ii)

2. Hello,
Originally Posted by chella182
The question goes:

Suppose the continuous random variable X has a probability density function (PDF)

$\displaystyle f_{X}(x)=\frac{c}{x^6}, x>1$

i) Show that for X to be a valid PDF, $\displaystyle c$ must be equal to $\displaystyle 5$.
ii) Calculate $\displaystyle E(X^6e^{-2X})$.

For part i) I keep getting $\displaystyle -5$, not $\displaystyle 5$ for $\displaystyle c$, and I just can't fathom out part ii)
i) Remember that a pdf has to be a positive function. Hence the answer you're given is more likely to be correct than yours
$\displaystyle 1=\int_1^{\infty} \frac{c}{x^6} ~dx=\int_1^\infty cx^{-6} ~dx$
so this is $\displaystyle \left. \frac{c}{-5} \cdot\frac{1}{x^6}\right|_1^{\infty}$

$\displaystyle =\lim_{x\to\infty} \frac{c}{-5}\cdot\frac{1}{x^6} {\color{red}-}\frac{c}{-5}\cdot 1=-\frac{c}{-5}=\frac c5$

just out from curiosity, which minus sign did you forget ?

ii) recall that for any continuous function, $\displaystyle \mathbb{E}(h(X))=\int_{\mathbb{R}} h(x)f(x) ~dx$, where f is the pdf of the random variable X.

This formula applies here and you get : $\displaystyle \int_1^\infty x^6e^{-2x} \cdot \frac{5}{x^6} ~dx$