If $\mathbb P[X_n=k]=C_n\max(k,n-k)$ for $k=1,...n-1$, then find $C_n$

Question

For an arbitrary $n\ge 3$, let $X_n$ be a random variable on $\{1,2,...,n-1\}$, whose distribution is

$\mathbb P[X_n=k]=C_n\max(k,n-k)$ for $k=1,...n-1$

What is the value of constant $C_n$ ?

I tried it, but obtained a strange result

So the necessary condition is $\displaystyle\sum\limits_{k=1}^{n-1}\mathbb P[X_n=k]=1$

I tried it for small numbers and i noticed that the summands are symmetric, It doesn't matter if $n$ is odd or even, the formula i got is for the sum without $C_n$

$\Big(2\displaystyle\sum\limits_{i=1}^{\lceil\frac{n-1}{2}\rceil}n-i\Big)-\Big(\frac{n-1}{2}-\lfloor{\frac{n-1}{2}}\rfloor\Big)n$

If $n$ is odd then the term after the sum vanishes.

And $C_n$ is then the reciprocal of this expression above.

Maybe we can simplify this ?

Answer 1

When $n$ is odd, $P\{X_n=k\}$ takes on decreasing values in the arithmetic progression $(n-1)C_n, (n-2)C_n, \ldots, \frac{n+1}{2}C_n$ as $k$ increases from $1$ to $\frac{n-1}{2}$, and increasing values in the arithmetic progression $\frac{n+1}{2}C_n, \frac{n+3}{2}C_n, \ldots, (n-2)C_n, (n-1)C_n$ as $k$ increases from $\frac{n+1}{2}$ to $n-1$. Pairing up the terms in the arithmetic progressions in the obvious way, we get $$\begin{align}\sum_{k=1}^{n-1} P\{X=k\} =1 &= \frac{n-1}{2}\left[(n-1) + \frac{n+1}{2}\right]C_n \\\Rightarrow \ C_n &= \frac{1}{3n^2-4n+1} = \frac{4}{(n-1)(3n-1)} = \frac{1}{\left(\frac{n-1}{2}\right)\left(\frac{3n-1}{2}\right)}\end{align}$$ where the last expression is of the form $\frac{1}{K}$ as desired by the OP.

When $n$ is even, we get similar arithmetic progressions but with a central term $\frac{n}{2}C_n$ that can be thought of as belonging to both arithmetic progressions. This gives $$\begin{align}\sum_{k=1}^{n-1} P\{X=k\} = 1 &= \frac{n}{2}\left[(n-1) + \frac{n}{2}\right]C_n - \frac{n}{2}C_n\\ \Rightarrow C_n &= \frac{4}{3n^2-4n} = \frac{4}{n(3n-4)} = \frac{1}{\left(\frac n2\right) \left(\frac{3n}{2} - 2\right)} \end{align}$$ once again expressing the result as $\frac{1}{K}$ for some integer $K$.

The formulas for odd and even $n$ can be combined into one expression if that is preferable, e.g.

$$\displaystyle C_n = \frac{1}{\left\lfloor \frac{3n^2-4n+1}{4}\right\rfloor} = \frac{1}{\left\lfloor \frac{n}{2}\right\rfloor \cdot \left\lfloor \frac{3n-2-(-1)^n}{2}\right\rfloor}, n \geq 3.$$

Answer 2

We have that $k \ge n-k$ iff $$k\ge \frac n2$$So, the pmf of $X_n$ can be written as $$P(X_n=k)=\begin{cases}C_n(n-k), & 1\le k< \frac n2 \\C_nk, & \frac n2\le k \le n-1 \end{cases}$$ But, we have that $$1 \le k < \frac n2 \implies -\frac n2 < -k \le -1$$ which gives that $$n-\frac n2 < n-k \le n-1$$Thus, $$\sum_{k=1}^{\lfloor \frac n2 \rfloor}(n-k)=\sum_{\lceil \frac n2 \rceil}^{n-1}k$$ Now, we can select cases depending on whether $n$ is even or odd.

1. $n$ is even. In that case $$\lfloor \frac n2 \rfloor=\lceil \frac n2 \rceil$$ and therefore $$\begin{align*}1&=\sum_{k=1}^{n}P(X_n=k)=C_n\left[\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}(n-k)+\sum_{k=\lceil\frac{n}{2}\rceil}^{n-1}k\right]=C_n\left[2\sum_{k=\frac{n}{2}}^{n-1}k-\frac n2\right]=\\&=C_n\left[2\cdot\frac{n-1+\frac{n}{2}}{2}\cdot(n-\frac{n}{2})-\frac n2\right]=C_n\left(\frac{3n}{2}-2\right)\left(\frac n2\right)\end{align*}$$ where for the last step I used the formula for summing up consecutive integers (hopefully without making a mistake). The above expression gives $$C_n=\frac{4}{(3n-4)n}$$
2. $n$ is odd. In that case $$\lfloor \frac n2 \rfloor<\frac n2 <\lceil \frac n2 \rceil$$ and therefore $$\begin{align*}1&=\sum_{k=1}^{n}P(X_n=k)=C_n\left[\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}(n-k)+\sum_{k=\lceil\frac{n}{2}\rceil}^{n-1}k\right]=C_n\left[2\sum_{k=\frac{n+1}{2}}^{n-1}k\right]=\\&=C_n\left[2\cdot\frac{n-1+\frac{n+1}{2}}{2}\cdot\frac{n-1}{2}\right]=C_n\left(\frac{3n-1}{2}\right)\left(\frac {n-1}{2}\right)\end{align*}$$ where for the last step I used the formula for summing up consecutive integers (hopefully without making a mistake). The above expression gives $$C_n=\frac{4}{(3n-1)(n-1)}$$