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I have this distribution function of a random variable X:

enter image description here

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 !

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  • $\begingroup$ 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 $C$? $\endgroup$ – whuber Jun 10 '16 at 13:43
  • $\begingroup$ 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. $\endgroup$ – user3275222 Jun 10 '16 at 13:46
  • $\begingroup$ And? What do you get when you plug $C$ into the formula? $\endgroup$ – whuber Jun 10 '16 at 13:48
  • $\begingroup$ I get that C=3/5, and then my integral gives 17/30, using both ways, mine and yours. :-) $\endgroup$ – user3275222 Jun 10 '16 at 13:54
  • $\begingroup$ 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. $\endgroup$ – whuber Jun 10 '16 at 14:22
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Edit: This answer is wrong. It does work out that way to 0.56667. I made an error with fractions.

Check you integral. It is not the problem.

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

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  • $\begingroup$ I re-checked it, with Maple, and it gives me 0.56667. $\endgroup$ – user3275222 Jun 10 '16 at 13:43
  • $\begingroup$ Yup. I mis-calculate. You are 100% correct. I work it out to 0.566667 too. Sorry. $\endgroup$ – Billyziege Jun 10 '16 at 13:57

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