Self-normalized weighted sum of random variables Given $n$ i.i.d. random variables $X_1, \dots, X_n$ with $X_i \sim \mathcal{N}(0,1)$  and weights $a_1, \dots, a_n \in [-1, 1]$ such that
$$
Y = \frac{P}{Q} = \frac{\sum_{i = 1}^{n} a_i X_i }{\sum_{i = 1}^{n} X_i}
$$
Thus, the numerator is $P = \sum_{i = 1}^{n} a_i X_i \sim \mathcal{N}(0,\sum_{i = 1}^{n} a_i^2)$ and $Q = \sum_{i = 1}^{n} X_i \sim \mathcal{N}(0,n)$.
I like to learn more background on Y; in particular, I'm interested in some closed-form approximation or bounds of Y based on the weights. . However, I don't know how to handle the fraction. I have found few papers which call $Y$ a self-normalized weighted sum; however, this seems to be not a common terminology as self-normalization typically refers to $\sum_{i = 1}^{n} X_i / \sum_{i = 1}^n X_i^2$.
 A: Because $(P,Q)$ is a linear transformation of $(X_1,\ldots, X_n),$ it has a binormal distribution.  Its mean is $(0,0)$ and its covariance matrix is determined from the bilinearity of covariance,
$$\operatorname{Cov}(P,Q) = \pmatrix{\sum a_i^2 & \sum a_i \\ \sum a_i & n}.$$
Let's regress $P$ against $Q:$ that is, let's find a variable $Z$ where $P = \beta Q + Z$ and $Z$ is uncorrelated with $Q.$  This gives the equation
$$0 = \operatorname{Cov}(Z, Q) = \operatorname{Cov}(P-\beta Q, Q) = \sum a_i - \beta n$$
with unique solution
$$\beta = \frac{1}{n}\sum_{i=1}^n a_i = \bar a,$$
whence
$$Z = P - \beta Q = P - \bar a Q,$$
implying
$$\operatorname{Var}(Z) = \operatorname{Var}(P - \bar a Q) = \sum_{i=1}^n a_i^2 - 2\bar a n + (\bar a)^2 n = n \operatorname{Var}(a).$$
Use this to rewrite the fraction in the form
$$\frac{P}{Q} = \beta + \frac{Z}{Q}.$$
The fraction obviously is a multiple of the ratio of uncorrelated (and therefore independent) standard Normal variables with the multiple equal to the standard deviation of the $a_i.$  It is well known (and easy to show that the ratio of independent standard Normals has a Cauchy distribution (aka Student's t with 1 degree of freedom).

This figure is a histogram of 100,000 draws of $Y=P/Q$ where $n=15,$ $\bar a = 0.52,$ and $\operatorname{SD}(a) = 0.31.$  (Approximately 4% of the values drawn would not fit on the horizontal axis.  As a result, the plotted densities of the histogram and red curve are all about 4% too large; but that's ok for these comparisons.)  Over it I have drawn the shifted, scaled Cauchy distribution in red.  For comparison, the graph of a standard Cauchy distribution is shown as a dotted black curve.  The agreement of the red curve with the histogram suggests this solution is correct.
This is the R code to produce the figure.
n <- 15
#
# Generate a random set of weights.
#
set.seed(17)
a <- runif(n, 0, 1) # (Limiting to positive values will make beta clearly nonzero)
#
# Generate many values of Y.
#
n.sim <- 1e5
X <- matrix(rnorm(n*n.sim), n.sim)
P <- X %*% a
Q <- X %*% rep(1, n)
Y <- P/Q
#
# Plot the distribution of those values as a histogram.
#
i <- abs(Y) <= 5
Y <- Y[i]
hist(Y, freq=FALSE, col="#f0f0f0", breaks=100)
#
# Compare the histogram to the theoretical solution.
#
beta <- mean(a)                   # The mean of the weights
sigma <- sqrt(mean((a - beta)^2)) # The SD of the weights
curve(dt((x-beta)/sigma, 1)/sigma/mean(i), n=1001, add=TRUE, lwd=2, col="Red")
abline(v=beta, lwd=2) # Mark the center of this distribution
#
# Draw a Cauchy PDF for reference.
#
curve(dt(x, 1), n=1001, add=TRUE, lwd=1,lty=3)

