Comparing Chance of Observing Event w/ Gaussian Variables

[daffyduck]

I am not sure how to do the following homework question:

Suppose X1, X2, ... are independent Gaussian variables with mean zero and variance 1. Consider the event that

X1 + X2 + ... + X2n ≥ 2na wth a > 0

Compare the chance of observing this event in the following two ways:
(i) by getting that X1 + X2 + ... + Xn ≥ na and Xn+1 + Xn+2 + ... +X2n ≥ 2na

(ii) by getting that X1 + X2 + ... + Xn ≥ 2na and Xn+1 + Xn+2 + ... + X2n ≥ 0

I tried letting Y1 = X1 + ... + Xn and Y2 = Xn+1 + ... + X2n.

For (i), Y1 and Y2 are each normally distributed with mean 0 and variance n,
so we have P(Y1 > an)P(Y2 > an) = P(Y1 > an)^2.

For (ii), P(Y1 > 2an)(1/2).

 

[Ray Vickson]

I am not sure how to do the following homework question:

Suppose X1, X2, ... are independent Gaussian variables with mean zero and variance 1. Consider the event that

X1 + X2 + ... + X2n ≥ 2na wth a > 0

Compare the chance of observing this event in the following two ways:
(i) by getting that X1 + X2 + ... + Xn ≥ na and Xn+1 + Xn+2 + ... +X2n ≥ 2na

(ii) by getting that X1 + X2 + ... + Xn ≥ 2na and Xn+1 + Xn+2 + ... + X2n ≥ 0

I tried letting Y1 = X1 + ... + Xn and Y2 = Xn+1 + ... + X2n.

For (i), Y1 and Y2 are each normally distributed with mean 0 and variance n,
so we have P(Y1 > an)P(Y2 > an) = P(Y1 > an)^2.

For (ii), P(Y1 > 2an)(1/2).

What is the problem? Both methods give wrong answers, just different wrong answers.

RGV

 

[daffyduck]

I need to know which approach is closer to the answer. They are both special cases of the actual event

 

[daffyduck]

The integrals u get for (i) and (ii) are hard to compare