Restrictions on a bounded variable's mean and standard deviation Given a random variable whose values are between $0$ and $1$, it is not difficult to see that the mean, or the expected value, is between $0$ and $1$, and the standard deviation is between $0$ and $\tfrac{1}{2}$.
However, not every combination is possible. If $\sigma=0$ then the mean could indeed be everywhere in $[0,1]$, but I tend to believe that $\sigma=\frac{1}{2}$ is only possible if $E[X] = \frac{1}{2}$.
Am I right? Are there simple restrictions involving a bounded variable's expected value and standard deviation?
Edit: Using the equality $\sigma(X) = \sigma(X-\frac{1}{2})$ we can get the equality $\sigma^2+(E[X]-\frac{1}{2})^2 = E[(X-\frac{1}{2})^2]$, the last expression being bounded by $0$ and $\frac{1}{4}$, so we get that the possible combinations must lie in the upper half circle (in the $\sigma-\mu$ plane) of radius $\frac{1}{2}$ around $\sigma=0,\mu=\frac{1}{2}$. Can we get a better result, or can any combination in the circle be reached?
 A: Tim Jones's answer is indeed the same as you have found.  
An alternative approach would be to note that for any given mean $\mu$ the variance $\sigma^2$ and standard deviation $\sigma$ is maximised when the random variable only takes the values $0$ or $1$, i.e. a Bernoulli random variable which takes the value $1$ with probability $\mu$ and the value $0$ with probability $1-\mu$.  In such a case the mean is $\mu$ and the standard deviation $\sigma = \sqrt{\mu(1-\mu)}$.  
Any smaller non-negative standard deviation can be achieved, for example when the random variable takes the value $1$ with probability $k\mu$, the value $0$ with probability $k(1-\mu)$ and the value $\mu$ with probability $1-k$, giving a mean of $\mu$ and a standard deviation $\sigma = \sqrt{k\mu(1-\mu)}$ with $0\le k\le 1$. So in general (corresponding to Tim Jones's result) $$0 \le \sigma \le  \sqrt{\mu(1-\mu)}.$$
This is indeed a semi-circle, as it can be rewritten by completing the square as $\sigma \ge 0$ and $$\sigma^2 +\left(\mu-\frac12\right)^2 \le \left(\frac12\right)^2$$
i.e. as you say "of radius $\frac{1}{2}$ around $\sigma=0,\mu=\frac{1}{2}$".
A: It's a lot easier to do this in terms of variance, but yes, using the fact that $X \in [0,1]$, we can get some sort of a bound on the Variance in terms of the Expectation;
\begin{align*}
Var(X) &= \mathbb{E}(X^2)-\mathbb{E}(X)^2\\
&= \sum x^2p(x) - \left(\sum xp(x)\right)^2
\end{align*}
So, we can manipulate this however we like - one such bound can be found thus
\begin{align*}
\sum x^2p(x) - \left(\sum xp(x)\right)^2 &\leqslant \sum xp(x) - \left(\sum xp(x)\right)^2\\
&= \mathbb{E}(X)(1-\mathbb{E}(X))\\
\implies Var(X) &\leqslant \mathbb{E}(X)(1-\mathbb{E}(X))
\end{align*}
The same sort of trick will work in the continuous case using integrals and densities rather than sums and masses, so, yes, it's only possible for a standard deviation of $\frac{1}{2}$ (Variance of $\frac{1}{4}$) to exist where the expectation is $\frac{1}{2}$, by maximising that bound, and no, not every combination is possible. If I were to choose an expectation and variance of $1\over 4$, I'd find no random variables in $[0,1]$ satisfying this.
