# Symmetricity of binomial distribution

For each random variable, $$X$$, define: $$Sym_X(c) = \frac{\Pr[X \geq \mu_X + c]}{\Pr[X \leq \mu_X - c]}$$.

I use this definition to measure how symmetric the distribution is.

Let $$X$$ be a binomial random variable, $$X \stackrel{}{\sim} Bin(n,p)$$.

1. As a rule of thumb, I know that when $$n$$ is large enough, the binomial distribution is "more" symmetric
2. What is $$Sym_X(c)$$? (For $$X \stackrel{}{\sim} Bin(n, p)$$) As an expression of $$n, p, c$$.
3. Let $$\sigma_X$$ be the STD of $$X$$. If $$c \leq \frac{1}{100}\sigma_X$$ can we bound $$Sym_X(c)$$ by a constant which does not depend on $$n,p,c$$?
• When $p$ and $c$ are fixed and $n \to \infty$, the central limit theorem implies $\text{Sym}_X(c) \to 1$, which formalizes your "#1". For #2, you can write the probabilities as sums, but I am not sure if this exact expression can be further simplified nicely. For #3, if $c$ is small, I feel like the numerator and denominator will both be close to $1/2$ since the median is close to the mean. Formalizing this intuition would take more work. – angryavian Feb 25 at 18:12
• First, thanks. Second, about #3, this is also my feel, I need to show that if $c$ is smaller than some constant times the STD, then the symmetricity will "win" as $n$ grows larger (actually, I can assume that $n \geq \frac{1}{p^2}$, so $n$ is large with respect to $\frac{1}{p}$). My goal is to prove that $Sym_X(C)$ can be bounded by a constant. – Eyal Yeh Feb 25 at 18:21