# So I have a probability density function 2t(0&lt=t&lt=1) &amp 0(t&lt0

So I have a probability density function 2t(0<=t<=1) & 0(t<0) & 1(t>0).I am just verifying whether I did this correctly. So the e(x) expected value is integral of 2t*t dt from 0 to 1 (the other intervals are all 0). This should be 2/3, correct? And then with 2 random independent variables of the same function, it would basically be double, which is 4/3. I.e. if you have Y = X1 + X2 and X1 and X2 are both the same function. I took the inverse transform method of this function which is basically integrated 2t, which is t^2, and then the inverse is X = sqrt(t); X2 = sqrt(t); etc.. I am generating proper values like the average e(x) over 10000 iterations turns into 1.3333 which is 4/3. If I just use X1, it is basically 2/3 or 0.66667. So I know my simulation is correct. However the probability that it is between 0.7 and 1.5? I am confused about that. I did integral of 2t between 0.7 and 1.0 and got 0.51 which is correct. However in simulation I added X2 and it went up to 0.61. How do I get that 0.10 difference in the probability P(0.7