To show that the Q function has the property Q(−x)=1−Q(x), we can use the definition of the Q function. The Q function is defined as Q(x) = P(X > x), where X is a standard normal distribution.
Let's consider Q(-x). According to the definition, Q(-x) = P(X > -x). Since the standard normal distribution is symmetric about the mean, we can rewrite this as Q(-x) = P(X < x).
Now, we can use the fact that the total probability under the normal distribution is equal to 1. Therefore, P(X < x) + P(X > x) = 1.
Substituting this in the previous expression, we get Q(-x) = P(X < x) = 1 - P(X > x) = 1 - Q(x).
Therefore, we have proved that Q(−x) = 1 − Q(x).
(b) Now, let's use the Chernoff bound to prove that Q(x) ≤ e^(2−x^2) for all x ≥ 0.
The Chernoff bound states that for any random variable X with moment generating function M(t), and for any positive t, we have P(X > t) ≤ e^(-ts) * M(t), where s is any positive constant.
In our case, X is a standard normal distribution, which has a moment generating function of M(t) = e^(t^2/2).
Using the definition of Q(x) = P(X > x), we can rewrite it as Q(x) ≤ e^(-tx) * M(t).
Substituting the moment generating function and rearranging the inequality, we get Q(x) ≤ e^(-tx) * e^(t^2/2).
To find the maximum value of the right-hand side, we can differentiate it with respect to t and set it equal to 0.
Differentiating e^(-tx) * e^(t^2/2) with respect to t, we get -xe^(-tx) * e^(t^2/2) + te^(-tx) * e^(t^2/2) = 0.
Simplifying this equation, we get -xe^(-tx) + te^(-tx) = 0.
Dividing both sides by e^(-tx), we have -x + t = 0.
Solving for t, we get t = x.
Substituting this value of t back into the inequality, we get Q(x) ≤ e^(-tx) * e^(t^2/2) = e^(-x^2/2).
Since e^(-x^2/2) is always less than or equal to e^(2−x^2) for x ≥ 0, we can conclude that Q(x) ≤ e^(2−x^2) for all x ≥ 0.
In conclusion, we have proven that Q(−x) = 1 − Q(x) and that Q(x) ≤ e^(2−x^2) for all x ≥ 0 using the properties of the Q function and the Chernoff bound, respectively.