Assume $X_1,X_2,X_3$ are IID $N(0,1)$ random variables. I want to show that $$P(X_1>0,X_1+X_2+X_3<0)>P(X_1+X_2>0,X_1+X_2+X_3<0)$$ I know I can show this by calcultating these two proababilities. But is there a way to prove this inequality without any evaluation?
Furthermore, I want to prove that $$P\left(\sum_{i=1}^{k}X_i>0,\sum_{i=1}^{n}X_i<0\right)>P\left(\sum_{i=1}^{m}X_i>0,\sum_{i=1}^{n}X_i<0\right)$$ for any $1\leq k<m\leq n$. Is this conjecture correct or not? If yes, how to prove it?
On
First of all we rewrite the first probability as follows $$ P(X_1 > 0,\ X_1 + X_2 + X_3 < 0) = P(X > 0, X + Z < 0) \\ = P(0 < X < Z) = P(X < Z\ |\ X,Z>0)/4= P(|X| < |Z|)/4$$ where $X=X_1$ and $Z=X_2+X_3\sim\mathcal{N}(0,2)$ are independent. We can write the second inequality using the same techniques to the following form $$ P(X_1 + X_2 > 0,\ X_1 + X_2 + X_3 < 0) = P(|W| < |Y|)/4$$ where $W = X_1 + X_2\sim\mathcal{N}(0,2)$ and $Y=X_3$. Note that we implicitly used the symmetry of the Gaussian distribution. Moreover, we used also for example that $P(X_1>0)=P(Z>0)=1/2$. It is vital that you can verify these equations for your understanding of the problem.
Now, let's tackle the problem from an intuitive point of view. The first probability essentially means that we are looking for a scenario where the random variable $X_1$ with variance $1$ is less away from its mean than $Z$ which has variance $2$. I think, you first need to convince yourself of this fact before you continue.
If you did that, we can formally complete the proof by noting that $Z/\sqrt{2},W/\sqrt{2}\sim \mathcal{N}(0,1)$. Hence, we could rewrite the second probability, for example, to $$P(|Z| < |X|)/4 = P(|Z|/\sqrt{2} < |X|/\sqrt{2})/4 = P(|A| < |B|/\sqrt{2})/4$$ where $A,B\sim \mathcal{N}(0,1)$ are i.i.d. Now, we rewrite the first probability now to the following form $$ P(|X| < |Z|)/4 = P(|X| < |Z/\sqrt{2}|\cdot\sqrt{2})/4 = P(|A| < |B|\cdot\sqrt{2})/4 $$ with the same distributions for $A$ and $B$. Hence, $$ P(X_1 > 0,\ X_1 + X_2 + X_3 < 0) = P(|A| < |B|\cdot \sqrt{2})/4 > P(|A| < |B|/\sqrt{2})/4 = P(X_1 + X_2 > 0,\ X_1 + X_2 + X_3 < 0). $$ Finally, note that this proof extends to your generalisation with $1\leq k < m\leq n$!
If anything is not clear in the above, feel free to ask!
Another way to prove this is to bring in geometric thinking. However, warning: the proof is no longer "elementary" because it uses a stronger theorem. Personally, I like the proof by @StanTendijck because it's elementary. OTOH, the following proof has the benefit of actually evaluating the probabilities (without integrals).
Here is the stronger theorem: the 3-D random vector $X = [X_1, X_2, X_3]$ is spherically symmetric about the origin. (It is obvious that the view along any of the 3 principle axes looks the same. It is much less obvious that the view along any radius is also the same.)
Obviously, $X_1 > 0$ is half the 3-D space, call it $U$. But similarly, $X_1 + X_2 + X_3 < 0$ is also half the 3-D space, call it $V$. Due to spherical symmetry, we dont have to consider the whole 3-D space and can instead can consider a sphere $S$ (of any fixed radius) and
$$P(X_1 > 0, X_1 + X_2 + X_3 < 0) = Volume(U\cap V \cap S) / Volume(S)$$
Now we have $X_1 = X \cdot [1, 0, 0]^T > 0$, and, $- X_1 - X_2 - X_3 = X \cdot [-1, -1, -1]^T > 0$. So the two hemispheres $U\cap S$ and $V \cap S$ are defined respectively by the vectors $u=[1, 0, 0]$ and $v=[-1, -1, -1]$ (the latter is often normalized but doesn't matter here). If two such vectors are angle $\theta$ apart (radians), then the intersection of the two hemispheres occupy a fraction ${\pi - \theta \over 2 \pi}$ of the whole sphere. This is obvious when you consider the plane defined by the two vectors.
So for these specific $u, v$ we have:
$$\cos(\theta) = {u \cdot v \over |u| |v|} = {-1 \over \sqrt{3}} \implies \theta \approx 2.19$$
$$P(X_1 > 0, X_1 + X_2 + X_3 < 0) = {\pi - \theta \over 2 \pi} \approx 0.152$$
Similarly, $X_1 + X_2 > 0$ is defined by the vector $w = [1, 1, 0]$ and now we need the angle $\phi$ between $w$ and $v$:
$$\cos(\phi) = {w \cdot v \over |w| |v|} = {-2 \over \sqrt{6}} \implies \phi \approx 2.53$$
$$P(X_1 + X_2 > 0, X_1 + X_2 + X_3 < 0) = {\pi - \phi \over 2 \pi} \approx 0.098$$
This also generalizes to higher dimensions, where:
$$P\left(\sum_{i=1}^{k}X_i>0,\sum_{i=1}^{n}X_i<0\right) = {\pi - A \over 2 \pi}$$
$$\cos(A) = {-k \over \sqrt{kn}} = -\sqrt{{k \over n}}$$
And your inequality is obtained when realizing
$$k \rightarrow n \implies \cos(A) \rightarrow -1 \implies A \rightarrow \pi \implies P\left(\sum_{i=1}^{k}X_i>0,\sum_{i=1}^{n}X_i<0\right) \rightarrow 0$$