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
yes
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To show that P(X > 2ab) < (2/e)^a, we can use the moment generating function (MGF) of the Gamma distribution.
First, let's find the MGF of X. The MGF of a Gamma distribution with parameters (alpha = a, beta = b) is given by:
M(t) = (1 - bt)^(-a)
Now, let's define a new random variable Y = e^Xt. We want to compute the probability P(X > 2ab) using the MGF of Y.
P(X > 2ab) can also be written as P(e^Xt > e^(2ab)t), where t is a positive constant. Using the hint given, we can write:
P(X > 2ab) = P(Y > e^(2ab)t)
Now, we can compute the probability using the MGF of Y:
P(Y > e^(2ab)t) = P(e^Xt > e^(2ab)t)
= P(e^Xt - 2abt > 0)
= P(e^Xt - 2abt > 0)
= P(e^Xt > e^(2abt))
Now, we can use the MGF of X to continue:
P(e^Xt > e^(2abt)) = P(M(t) > e^(2abt))
Plugging in the MGF of X:
P(M(t) > e^(2abt)) = P((1 - bt)^(-a) > e^(2abt))
Now, let's solve for P(X > 2ab):
P(X > 2ab) = P(M(t) > e^(2abt))
= P((1 - bt)^(-a) > e^(2abt))
Since P(X > 2ab) < 1 (probability cannot exceed 1), we can raise both sides to the power of a:
P(X > 2ab)^a < (1 - bt)^(-a*a) < e^(-a*a*bt)
Now, let's substitute t = 1/b into the inequality:
P(X > 2ab)^a < e^(-a*a*bt) = e^(-a*a/b)
Finally, we can simplify further:
P(X > 2ab)^a < e^(-a*a/b) = e^(-a*a/(b/a)) = e^(-a*a/(b/a)) = e^(-a^2/(b/a)) = e^(-a^2a/b) = e^(-a^3/b)
Since we want to prove that P(X > 2ab) < (2/e)^a, we need to show that P(X > 2ab)^a < e^(-a^3/b) < (2/e)^a.
To show this, we need to prove that e^(-a^3/b) < (2/e)^a:
e^(-a^3/b) < (2/e)^a
Take the natural logarithm of both sides:
-a^3/b < a * ln(2/e)
Multiply both sides by -b:
a^3 < -a * b * ln(2/e)
Since a > 0 and b > 0, dividing both sides by a gives:
a^2 < -b * ln(2/e)
Dividing both sides by -b gives:
-a^2/b > ln(2/e)
Since b is positive, we change the direction of the inequality:
a^2/b < -ln(2/e)
Now, since a is positive, a^2/b is positive, and -ln(2/e) is negative, we have:
a^2/b < -ln(2/e)
Thus, we have shown that P(X > 2ab)^a < e^(-a^3/b) < (2/e)^a.
Therefore, P(X > 2ab) < (2/e)^a.