Finding the expected value of a continuous random varibale when the commulative distribution is given

I have this distribution function of a random variable X:

I wish to find E(X).

I have used derivatives to get the density function, compared it to 1, and found that f(t) = (4/5)t+(3/5). I then used integral of tf(t) over the range of 0 to 1, and got 0.56667. According to the answer I got, it's incorrect (maybe the answer is incorrect, not me?). Can you please assist? Thank you !

• Because the expectation of any positive variable is the integral of its survival function, $$\mathbb{E}(X)=\int_0^\infty(1-F(t))dt=\int_0^1(1-Ct-2t^2/5)dt=1-C/2-2/15.$$ Thus, you need only compute $C$. Could you show us how you found $C$? – whuber Jun 10 '16 at 13:43
• I wasn't planning to use survival functions for this. What I did was simple. I found the density using the derivative of the F, then I compared the density to 1 over the entire range of 0 to 1. This gave me C. – user3275222 Jun 10 '16 at 13:46
• And? What do you get when you plug $C$ into the formula? – whuber Jun 10 '16 at 13:48
• I get that C=3/5, and then my integral gives 17/30, using both ways, mine and yours. :-) – user3275222 Jun 10 '16 at 13:54
• Okay. But let's notice something: $C$ is not uniquely defined. Any value of $C$ between $0$ and $3/5$ is legitimate, because the constraints on it (namely, that $F(t)$ lie between $0$ and $1$ for $0\le t\le 1$ and $F$ is non-decreasing in that same interval) only imply that $C \le 3/5$ and $C\ge 0$, respectively. Therefore, any expectation between $1-0/2-2/15/=26/30$ and $1-3/10-2/15=17/30$ could be "correct." The best answer would be to give this entire range of results. – whuber Jun 10 '16 at 14:22

Your approach is correct. Your probability density is correct, but your $$\int_0^1 t f(t) dt \ne 0.56667$$. Rework this integral.