Probability of sum of gaussian random variables

Question

We are given $n$ independent Gaussian random variables $x_i \sim N(0,1)$ and contants $c_i \geq 0$. Let the set $K$ contain $k$ indices corresponding to the smallest $c_i$. What is the probability that $\sum_{i \in K} c_ix_i \leq 0$.

Ok, so what I did so far is the following: Since $x_i \sim N(0,1)$ we obtain the random variable $c_ix_i \sim N(0, c_i^2)$ and moreover $\sum_{i \in K} c_ix_i \sim N(0, \sum_{i \in K} c_i^2$). Hence $\sum_{i \in K} c_ix_i$ is a random normal distributed variable and we can compute $Pr(\sum_{i \in K} c_ix_i \leq 0) = \frac{1}{\sqrt{2\pi \sum_{i \in K} c_i^2}} \int_{-\infty}^0 e^{-\frac{1}{2}\left(\frac{t}{\sqrt{\sum_{i \in K} c_i^2}}\right)^2}dt=\frac{1}{2}$

I am sceptic if this can truly be the right answer. The set $K$ is not chosen randomly but be explicitely select the indices corresponding to the smallest $c_i$, hence should the solution not depend on $c$?

Answer 1

Since the coefficients are not random $\sum_{i \in K} c_ix_i$ has normal distribution with mean $0$ (unless each $c_i$ is $0$) . Hence the probability is $\frac 1 2$.

Answer 2

Given $X_i \sim \mathcal{N}(0,1)$ i.i.d. r.v.s, we have $c_iX_i \sim \mathcal{N}(0,c_i^2)$, for $i \in K$

Let's define the r.v. $X=\sum\limits_{i \in K}c_iX_i$

$\implies E[X]=\sum\limits_{i \in K}E[X_i]=0$ and $V[X]=\sum\limits_{i \in K}V[X_i]=\sum\limits_{i \in K}c_i^2\quad$ by independence.

Also, the M.G.F. (moment generating function) for $X$ is:

$M_X(t)=E[e^{tX}]=E\left[e^{t\sum\limits_{i \in K}c_iX_i}\right]=\prod\limits_{i \in K}E[e^{tc_iX_i}]=\prod\limits_{i \in K}e^{0.t+\frac{c_i^2t^2}{2}}=e^{0.t+\frac{\sum\limits_{i \in K}c_i^2t^2}{2}}$, by independence

$\implies X \sim \mathcal{N}(0,\sum\limits_{i \in K}c_i^2) \implies Pr(X \leq 0) = \frac{1}{2}$