Probability for sum of $n$ uniform $(0, 1)$ being larger than sum of $n+1$ uniform $(0,1)$

Question

For $$U_1, \cdots, U_n, W_1, \cdots, W_{n+1} \sim \text{DiscreteUniform}(0,1)$$ all i.i.d., I want to show: $$\mathbb{P}\left(\sum_{i=1}^n U_i \geq \sum_{i=1}^{n+1} W_i \right) = 0.5$$

I was trying to do this via induction, and conditioning $W_i$'s as it can only take values either 0 and 1. But, still, the equation was getting TOO messy and I could not proceed.

My intuition says, as Ws and Us has same PMFs, the only thing that affects to the probability calculation is $W_{n+1}$ which takes either 0 and 1 with equal probability.

Any easy proof for the assertion I made above?

P.S. I am adding probability plots for $n \in \{1, \cdots, 50\}$, where for each $n$, I executed 100000 Monte Carlo simulations (assuming this number is enough, I have not performed any standard error analysis).

Answer 1

Let $X = \sum\limits_{i=1}^{n} \left(1-U_i\right) +\sum\limits_{i=1}^{n+1} W_i = n-\sum\limits_{i=1}^{n}U_i + \sum\limits_{i=1}^{n+1} W_i$. It is clear that $X \sim Bin\left(2n+1, \frac12\right)$ and you are looking for the probability that $X \le n$. However, $X$ and $2n+1-X$ have the same distribution so

\begin{align} 2P\left(X\le n \right) &= P\left(X\le n\right) + P\left(2n+1-X\le n\right)\\ &= P\left(X\le n\right) + P\left(X\ge n+1\right) = 1. \end{align}

So the statement is true.

Answer 2

I think it was easier using the fact that sum of Bernoulli is Binomial. Considering that, here is my proof for the assertion:

Let $Z_i = 1 - U_i$ for all $i \in \{1, \cdots, n\}$. Now, we start by noting $U_i$ and $W_i$ are Bernoulli(0.5) i.i.d.. Then $Z_i \sim \text{Bernoulli}(0.5)$ for all $i$ and $Z_i$ and $W_i$ are also i.i.d.

Now, note that $$\mathbb{P}\left( \sum_{i=1}^n U_i \geq \sum_{i=1}^{n+1} W_i \right) = \mathbb{P}\left( \sum_{i=1}^n (1-Z_i) \geq \sum_{i=1}^{n+1} W_i \right) = \mathbb{P}\left( n \geq \sum_{i=1}^n Z_i + \sum_{i=1}^{n+1} W_i \right)$$

As sum of Bernoulli is Binomial, we have $$\sum_{i=1}^n Z_i + \sum_{i=1}^{n+1} W_i = Q \sim \text{DiscreteBinomial}(2n+1, 0.5)$$

Hence, $$\mathbb{P}\left( \sum_{i=1}^n U_i \geq \sum_{i=1}^{n+1} W_i \right) = \mathbb{P}\left( n \geq Q \right)$$

There is NO closed form for CDF binomial, though using R and testing for $n \in \{1, \cdots, 100000\}$, I consistently obtained 0.5. So I have no idea how to continue here.

Answer 3

Let $W$ and $U$ be iid.

Then by symmetry: $$P(U>W)+P(U\geq W)=P(U>W)+1-P(U<W)=1$$ Moreover let it be that: $$U>W\implies U\geq W+1$$

Then if $B\sim\text{Bernoulli}(0.5)$ and independent wrt $U$ and $V$:$$P(U\geq W+B)=\frac12P(U>W)+\frac12P(U\geq W)=\frac12$$

This can be applied on $U:=\sum_{i=1}^nU_i$ and $W:=\sum_{i=1}^nW_i$ and $B=W_{n+1}$.