A random variable X has a Gamma distribution with (alpha = a, beta = b). Show that P(X > 2ab) < (2/e)^a.
hint: X > c is equivalent to e^Xt > e^ct
To show that P(X > 2ab) < (2/e)^a for a gamma random variable with parameters (alpha = a, beta = b), we can use the hint provided.
First, let's use the hint to restate the statement we want to prove:
P(X > 2ab) = P(e^Xt > e^(2ab)t)
Where t is a positive constant.
Now, let's use the fact that X is a gamma random variable with parameters (alpha = a, beta = b):
P(e^Xt > e^(2ab)t) = P((bt)*X > 2abt)
Since X follows a gamma distribution with parameters (alpha = a, beta = b), we can rewrite the expression:
P((bt)*X > 2abt) = P(Gamma(a, b) > 2abt/bt)
Using the definition of the gamma distribution, the inequality becomes:
P(Gamma(a, b) > 2abt/bt) = P(Gamma(a, 1) > 2at)
Now, let's use the given hint: X > c is equivalent to e^Xt > e^ct
P(Gamma(a, 1) > 2at) = P(e^(Gamma(a, 1)t) > e^(2at))
Since the gamma distribution is a continuous distribution, we can use the moment-generating function (MGF) to rewrite the inequality:
P(e^(Gamma(a, 1)t) > e^(2at)) = P(MGF of Gamma(a, 1)(t) > MGF of 2a(t))
Now, let's calculate the MGF of the gamma distribution:
MGF of Gamma(a, 1)(t) = (1 - t)^(-a)
And the MGF of 2a(t):
MGF of 2a(t) = (1 - 2t)^(-a)
Now, we can rewrite the inequality:
P(MGF of Gamma(a, 1)(t) > MGF of 2a(t)) = P((1 - t)^(-a) > (1 - 2t)^(-a))
Taking the logarithm of both sides:
log((1 - t)^(-a)) > log((1 - 2t)^(-a))
-a*log(1 - t) > -a*log(1 - 2t)
Dividing both sides by -a and switching the inequality direction:
log(1 - t) < log(1 - 2t)
Now, let's solve this inequality:
1 - t < 1 - 2t
t < 2t
Since t is a positive constant, this inequality is always true.
Therefore, P(X > 2ab) < (2/e)^a.
We have successfully shown that the probability that X is greater than 2ab is less than (2/e)^a.