Suppose that a and b are real numbers such that the function defined by p(x)=a+bx2 for 0≤x≤1 and p(x)=0 for all other x
is the p.d.f. of some random variable Y. What is the smallest possible value of the mean of Y? Enter your answer as a fraction, such as 1/2.
Here's what I have so far. From the first axiom where the probability from 0 to 1 must equal 1, a+b/3=1.
Now the indefinite integral of the mean is x*p(x) from 0 to 1 is 1/4(2a+b). Now to find the minimum mean, i thought to differentiate but it's in terms of two variables. So I substituted a=1-b/3 into 1/4(2a+b) but the differentiation leads to 1/12 the constant. I'm on the right track with my boundaries of the values of a and b and needing to derive to find the minimum. I just mess up the step of deriving the 1/4(2a+b) using these boundaries.
Best Answer
Because $p(1)\geq 0$, we must have $b \geq -a$. With $a = 1 - {1\over 3}b$, this implies $b \geq -{3\over 2}$.
That is the constraint you seem to have missed. With this constraint, the expected value is ${1\over 2}a + {1\over 4}b = {1\over 2} + {1\over 12}b \geq {1\over 2} + {1\over 12}(-{3\over 2}) = {3\over 8}$.
(In other words, this is not a local minimum found by setting a derivative equal to zero.)