Let $X_1, X_2, \dots$ be a sequence of i.i.d. random variables with mean zero (e.g., $N(0,1)$). Let $n > m$ and $c \geq 0$. I want to show that
$$ P\left(\sum_{i=1}^n X_i \leq c \right) \leq P \left( \sum_{i=1}^m X_i \leq c \right).$$
In view of existing concentration bounds like Hoeffding's inequlity which scale with the the length of the sequence, i.e. $n$ and $m$, I would think that the above statement should hold.
Edit: Since it was pointed out that this doesn't hold for specific cases where $n$ and $m$ are small and the distribution of $X_i$ is discrete, assume that $X_1, X_2, \dots$ are Gaussian and $n$ sufficiently large.
Best Answer
From the gaussian assumption, we can calculate analytically the two probabilities, it suffices to notice that $\sum_{i=1}^nX_i\sim \sqrt{n}\mathcal{N}(0,1)$ then $$\mathbb{P}\left(\sum_{i=1}^nX_i\le c \right) = \mathbb{P}\left(\mathcal{N}(0,1)\le \frac{c}{\sqrt n} \right) = \Phi\left(\frac{c}{\sqrt n}\right)$$
As $\frac{c}{\sqrt n} <\frac{c}{\sqrt m}$ for $c \ge 0$ and $n >m$, we have $$\mathbb{P}\left(\sum_{i=1}^nX_i\le c \right)\le \mathbb{P}\left(\sum_{i=1}^mX_i\le c \right)$$