Mathematical Modelling Question

1. May 28, 2006

Hummingbird25

HELP: Mathematical Modelling Question

Hi

Given $$X_1 \ldots X_n$$ be stochastic independent variables with the distribution functions $$F_X_{1}, \ldots ,F_X_{n}$$. $$U = min(X_1 \ldots X_n)$$ and $$V = min(X_1 \ldots X_n)$$.

$$F_{U}$$ and $$F_{V}$$ for U and V, and let $$F_{U,V}$$ be simultaneously distribution functions for the stochastic vectors (U,V).

Then show that

$$F_{V} (s) = \Pi \limit_{i=1} ^{n} F_{X_i} (s)$$ where $$\forall s \in \mathbb{R}$$

I can see that if I expand the sum I get

$$F_X_{1}(s) + F_X_{2}(s) + F_X_{3}(s) + \ldots + F_X_{i}(s)$$ where $$1 \leq i \leq n$$

Doesn't that mean that

$$F_X_{1}(s) + F_X_{2}(s) + F_X_{3}(s) + \ldots + F_X_{i}(s) = (F_X_{1}(s) \ \mathrm{U} \ F_X_{2}(s) \ \mathrm{U} F_X_{3}(s) \ \mathrm{U} \ \ldots \ \mathrm{U} \ F_X_{n}(s))$$ ??

Since $$\sum_{i=1} ^{n} P(A_i) = P(A_1) + P(A_2) + P(A_3) + \ldots + P(A_n)$$

Sincerely
Hummingbird

Last edited: May 28, 2006
2. May 28, 2006

HallsofIvy

Staff Emeritus
May we assume that
$$V = min(X_1 \ldots X_n)$$
was actually supposed to be
$$V = max(X_1 \ldots X_n)$$

3. May 28, 2006

Hummingbird25

My assignment uses U and V to distingues between min and max, but I guess it doesn't make that a bit a difference in the final calculation.

Sincerely Humingbird