Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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]) ?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $0$.

share|improve this answer
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.