I am trying to find the expected value of a random variable which results in the following summation: $$ \frac{1}{n^{2n}} \sum_{r=1}^n r \binom{n}{r} (n-r)^{2n} $$ Can someone please help me solve it
Thanks in advance!!
P.S.: I am new on here and this is my first question.. please forgive for any mistakes made in asking questions or any other associated thing
Not sure if a closed form is possible but it is possible to get a good approximation as follows. First, we use the following identity which can be proved using the binomial theorem and taking derivatives.
$$ \sum_{r=1}^n rx^r \binom{n}r = nx(x+1)^{n-1}. $$
For an upper bound, note that your quantity can be written as $$ \sum_{r=1}^n r \binom{n}r \left( 1 - \frac{r}n \right)^{2n} \le \sum_{r=1}^n r \binom{n}r e^{-2r} = ne^{-2}(1+e^{-2})^{n-1} = O(1.14^n) $$ where we have used the estimate $1-x \le e^{-x}$ and the above identity.
A lower bound can be proved similarly. Note that $1-x \ge e^{-1.4x}$ for $x \le 0.5$. Hence, we can say that your quantity is at least $$ ne^{-2.8}(1+e^{-2.8})^{n-1} = O(1.06^n)$$ minus an error term that is of the order $$ \binom{n}{n/2} \frac{1}{4^n} = O(.5^n) $$ so it is negligible.