Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we let $X$ be a random variable with c.d.f. $F(X)=x^3$ for $0 \le x \le 1$ as my cumulative distribution function and I need to find $P(X\ge \frac12)$ is this correct? $$\int_{1/2}^1 x^3dx=\frac{15}{64} $$ or is the correct way to find the probability this way: $$1-\int_0^{\frac12} x^3dx=\frac{64}{65}$$ I think it is the second way because by looking at a graph it would make sense, however, it seems that this probability is too high for it to be correct.

share|cite|improve this question

You have $P(X\ge 1/2) = 1 - P(X\le 1/2) = 1 - 1/8 = 7/8.$ Don't integrate again. The density is $f_X(x) = 3x^2$, $0\le x \le 1$.

share|cite|improve this answer
So for a cumulative density function were simply "plugging in" the value we want into $F(X)$? – TheHopefulActuary Nov 7 '12 at 0:46
what makes you think $0<x<1$? – Alex Nov 7 '12 at 0:49
@ncmathsadist i appologize, the problem states $F(x)= x^3$ for $0 \le x \le 1$ – TheHopefulActuary Nov 7 '12 at 0:55

Just an addition to ncmathsadist's answer. Nothing in your question says about the support of the distribution, i.e. on which set $X$ is defined. Hence what you are doing, i.e. $\int_{\frac{1}{2}}^{1} F(x)dx$ does not make any sense, but what you know for sure is that $F(\infty)=1$ and $F \big(\frac{1}{2} \big)=\frac{1}{8}$ and this is all you need.

share|cite|improve this answer

Your Answer


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.