# Continuous random variable, density functions.

Let X be a continuous random variable with density fx(s) = k(s$^{-6}$) for s $\ge$ 1 and fx(s) = 0 for s<1.

1.) Determine the value of the constant k.

2.) Calculate E(X), VAR(X)

3.) Calculate P(X $\in$ [-2,-0.75])

I know I must integrate with respect to s and for E(X) = (X$^2$-E(X)) and VAR(X) = (X-E(X))$^2$, but how can I format it correctly to find E(X), VAR(X), k, and P(X $\in$ [-2,-0.75]) ?

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You want $$\int_1^\infty k s^{-6}\,ds=1.$$ Integrate. You should get $\frac{k}{5}$, so $k=5$.
For $E(X)$, use the usual formula $$E(X)=\int_{-\infty}^\infty sf_X(s)\,ds.$$ Our density function is $0$ except on $[1,\infty)$, so $$E(X)=\int_1^\infty (s)(5s^{-6})\,ds.$$ The integrand is $5s^{-5}$. Now do the integration.
For the variance, you can use $E((X-\mu)^2)$, where $\mu=E(X)$. However, as usual it will be easier if you use the fact that the variance is $E(X^2)-(E(X))^2$. So the only thing not yet known is $E(X^2)$. But $$E(X^2)=\int_1^\infty (s^2)(5s^{-6})\,ds.$$ Simplify the integrand, and integrate.
For the probability, if there is no typo, our interval is an interval of negative numbers. On the negatives, the density is $0$, so the integral of $f_X(s)$ over this interval is $0$.
@Q.matin: One could say, as you do, that the density function only applies above $1$. And indeed if we wanted $\Pr(2\le X\le 3)$ we would integrate $5s^{-6}$ from $2$ to $3$. For technical reasons, I prefer to think of the density function as being $0$ if $s\lt 0$. – André Nicolas Oct 2 '12 at 19:17