Monte Carlo VaR

mediumQuantacceptance —~25 min

Implement mc_var(S, mu, sigma, days, n, seed, alpha) — a Monte Carlo value-at-risk. Call random.seed(seed), then for each of n paths draw one random.gauss(0, 1) per day and step S·exp((mu - sigma²/2)·dt + sigma·sqrt(dt)·z) with dt = 1/252, for days steps. VaR is the loss S - S_T at the alpha quantile: sort the losses descending and take index int(alpha·n).

It is the risk desk's morning number, and the workbench's Monte Carlo VaR bot. Return VaR as a fraction of S rounded to 4 decimal places.

example 1

input

S=100.0, mu=0.05, sigma=0.2, days=1, n=2000, seed=7, alpha=0.05

output

0.0218

example 2

input

S=100.0, mu=0.05, sigma=0.2, days=10, n=2000, seed=7, alpha=0.05

output

0.0622

constraints

S > 0
1 ≤ days ≤ 60
100 ≤ n ≤ 20_000
0 < alpha < 0.5

teacher · 2 hints · the first is free

↗ read the theory

the chapter this problem is cut from

▶ open in the workbench

run the same math on a real tape

●solution.py· utf-8 · spaces:4

import random
from math import exp, sqrt


def mc_var(S: float, mu: float, sigma: float, days: int, n: int, seed: int, alpha: float) -> float:
    """Monte Carlo VaR as a fraction of S, rounded to 4 dp."""
    random.seed(seed)
    dt = 1 / 252
    losses = []
    for _ in range(n):
        # TODO: the horizon is `days` steps, not one
        z = random.gauss(0, 1)
        s = S * exp((mu - 0.5 * sigma ** 2) * dt + sigma * sqrt(dt) * z)
        losses.append(S - s)
    losses.sort(reverse=True)
    return round(losses[int(alpha * n)] / S, 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to mc_var()

S=100.0, mu=0.05, sigma=0.2, days=1, n=2000, seed=7, alpha=0.05

expected

0.0218

the judge calls mc_var() with exactly these inputs · outputs are compared to 6 decimal places

● standing bypython 3.13 in your browser · pyodide 0.28.3 · judged on the server to 6 dp⌘↵ run · ⌘⇧↵ submit

paper only · educational · not advice