$X$ is uniform, $Y$ is binomial. When is $Var(X)>Var(Y)$?

Question

Let $X$ be a discrete random variable, uniformly distributed over $\{1,2,..., n\}$. Let $Y$ be a binomial random variable. For what values of $n$ (in terms of $p$ is $var(X)>var(Y)$?

This is my solution so far:

$Var(X)=\frac{n^2-1}{12}$

$Var(Y)=np(1-p)$

$\frac{n^2-1}{12} > np(1-p)$

$n^2-1>12np(1-p)$

$n^2-12np(1-p)-1>0$

$n>\frac{12p(1-p)\pm \sqrt{(12p(1-p))^2-4(1)(-1)}}{2(1)}$

$=\frac{12p(1-p)\pm\sqrt{144p^2(1-p)^2+4}}{2}$

$=\frac{12p(1-p)\pm 2 \sqrt{36p^2(1-p)^2+1}}{2}$

$=6p(1-p)\pm \sqrt{36p^2(1-p)^2+1}$

Answer 1

Your calculation is almost correct.

The first error is when you isolate $n>\cdots $ . Actually you have a quadric with two roots ($r_1, r_2$) and the solutions are $n<r_1$ or $n>r_2$. However, the lower root is negative (you should show this) , so only the other one is relevant.

This is

$$ r_2 = 6 p\, \left( 1-p\right) +\sqrt{36 p^2 {{\left( 1-p\right) }^{2}}+1}$$

(I believe you messed up an exponent). Hence, it's true that, for any fixed $p$, you have infinely many $n$ for which the uniform has bigger variance.

Why does this make sense even though $Y$ can take on infinitely many values

That's false. Both have a finite range: the uniform takes $n$ values and the Binomial takes $n+1$ values.