Can the Expected Value of a Continuous Random Variable Take on any Value

Question

So given a continuous random variable $X$ with density function $a + bx^2$ for $0\le x \le 1$ and $0$ otherwise.

If any values can be assigned to $a$ and $b$, can the distribution take on any expected value?

My immediate thought were that yes because there would exist a combination of $a$ and $b$ that would yield any mean. I then remembered that the PDF must always be non-negative and that the CDF must equal $1$ as $X$ approaches infinity. This caused me to doubt my initial instincts and I now believe that there are likely values that the conditions for the CDF or PDF would not be met. I would be interested to hear your reasoning on the problem.

Answer 1

Clearly the expected value cannot be outside the range $[0,1]$, so if "any expected value" includes anything outside this the answer is no. Since this is a pdf, the integral from $-\infty$ to $+\infty$, which is the integral from $0$ to $1$ has to equal $1$. That gives you a relationship between $a$ and $b$ and you can find the expected value in terms of one of them. The pdf has to be nonnegative, so that gives some limits to $a,b$. See if it can range over $[0,1]$. I believe the answer is no.

Answer 2

For the given function be a PDF, the integral over $0$ to $1$ must be 1. This condition gives us that $$a+\frac{b}{3} = 1.$$ Now, $$E[X] = \int_{0}^{1}xf(x)dx.$$ This gives $$E[X] = \frac{a}{2}+\frac{b}{4}.$$ Substituting from previous equation we have,$$E[X] = \frac{1}{2}+\frac{b}{12}$$

Now we have to see what possible values can $b$ take to evaluate, all possible values for expectation. This can be done by seeing the possible values of $b$ where PDF is never negative.