Cumulative Distribution Function calculation I've got a question here that I've been working and I've ran into some difficulty.
"Let X be a mixed random variable with distribution function:
F(x) = 0 if x < 0
   cx + 1/4 if 0 <= x < 1 (c is some constant)

   1/3  if 1 <= x <2

   1    if x >= 2

As for my notation above, I'm unfamiliar with LaTex, but what I'm trying to show you is a piecewise function. Here <= means "less than or equal to".
Now the question asks to find the value of c, given the expected value is 1. Here is what I have done:
First, find the values of the probabilities at 0,1 and 2
Using limits, we find P(X=0) is 1/4.
Similiarly, we find P(X=1) is 1/12 -c (here I subtracted c+1/4 from 1/3, using the idea of limits again)
Finally, we find P(X=2) is 1- 1/3 = 2/3.
Now, use these values to calculate the expected value:
Here is where the confusion arises. First, I split the E[X] into three components:
1) the definite integral of d/dx [(cx + 1/4)times x] between 0 and 1
2) 1/12 - c times 1
3) 2 times 2/3 
Now, I add these all up and equate:
c/2 +4/3 +1/12 -c = 1,
hence solving c = 5/6, however this leads to a negative probability for P(X=1), as 1/12 - 5/6 is less than zero.
Where have I gone wrong in my above calculations? I think it's to do with my expected value calculation.
 A: Your calculations are all correct. Even with somewhat less work than you did, one can see that it is impossible for the mean to be $1$. Our random variable is non-negative, and is equal to $2$ with probability $2/3$, so has it has mean at least $4/3$.  Presumably there is a typo in the description of the cumulative distribution function. 
A: Problems involving mixed random variables are sometimes easier to solve using an
alternative definition of the expected value.  For a nonnegative
random variable, which is what you have, this alternative definition is
$$E[X] = \int_0^{\infty} [1 - F(x)]\,\mathrm dx$$
where the square brackets should not be removed and the
integral on the right expressed as the difference of two integrals from
$0$ to $\infty$.
For your CDF, you can write
$$E[X] = \int_0^1 \left[\frac{3}{4} - cx \right]\,\mathrm dx
+ \int_1^2 \frac{2}{3}\,\mathrm dx + \int_0^{\infty} 0 \,\mathrm dx$$
So, compute the value of the integrals which will give you
a function $g(c)$ of $c$, set $E[X] = g(c) = 1$, and solve for $c$.
Edit:  Ooops, as André's answer shows, you will get a nonsensical
answer if you follow the above procedure for your CDF, but the
idea will work for a correctly formulated problem.
