Large variance implies small probability of being in interval Let $X$ be some random variable on a real interval $[0,\infty)$. Let $[a,b]\subset [0,\infty)$. Intuitively, it seems reasonable that if the variance of $X$ denoted by $\sigma_X^2$ is large compared to the size of the subset $b-a$, then the probability of being inside the subset should be small (if it was high then the variance would be small)
I am thus looking for a bound of the type
$P(a\leq X\leq b)\leq f(\sigma_X,a,b)$
where $f$ is some function that decreases as $\sigma_X$ gets larger (and probably increases with $b-a$).
There are many bounds, as can be seen already on Wikipedia, but most of them show that if the variance is small then the probability of being far away from the mean is small (e.g. Chebyshev/Markov bound). I want the opposite: If the variance is large then the probability of being in any given (small) interval is small. I have unfortunately not manages to adapt those bounds to what I need (or if there is an existing one not recognized it as such)
 A: The reason why you cannot find a bound is because it doesn't exist.  You can make the variance arbitrarily large, but this doesn't force the probability of being in a pre-specified interval to tend to a nontrivial upper bound asymptotically unless the support is finite.
To see why, define for $a < b < c < d$ and $\theta \in (0,1)$ a random variable $$f_X(x) = \begin{cases} \frac{\theta}{b - a}, & a \le x \le b \\ \frac{1-\theta}{d-c} & c \le x \le d. \end{cases}$$ Then $$\operatorname{E}[X] = \frac{(a+b)\theta + (c+d)(1-\theta)}{2},$$ and $$\operatorname{Var}[X] =\frac{1}{12} \theta  \left(4 \left(a^2+a b+b^2\right)+2 \left(-3 (a+b) (c+d)+c^2+4 c
   d+d^2\right)\right)-\frac{1}{4} \theta ^2 (a+b-c-d)^2+\frac{1}{12} (c-d)^2.$$  In the special case where $a = b-1$ and $d = c+1$, we get $$\operatorname{Var}[X] = \frac{1}{12} + (c-b+1)^2 \theta(1-\theta),$$ and it becomes immediately apparent that this can be made arbitrarily large by increasing the difference $|c-b|$; yet $\Pr[b-1 \le X \le b]$ remains constant with respect to the parameter $\theta$; in turn, then, this probability can remain arbitrarily large.
