Question

Let \(X\) be a random variable with mgf \(M(t),-h

Short answer

The probability of \( X \) being greater than or equal to \( a \) is bounded by \( e^{-at} M(t) \) for \( 0 < t < h \) and the probability of \( X \) being less than or equal to \( a \) is also bounded by \( e^{-at} M(t) \) for \( -h < t < 0 \). These conclusions are derived from Theorem 1.10.2 with appropriate substitutions and understanding of moment generating functions.

Step-by-step solution

Formulate Using Theorem 1.10.2

From Theorem 1.10.2, we have for an increasing function \( u(x) \) and constant \( c \), \[ P(X > a) \leq \frac{E[u(X)]}{u(c)} \] Substituting \( u(x) = e^{tx} \) and \( c = e^{ta} \), we get \[ P(X \geq a) \leq \frac{E[e^{tX}]}{e^{ta}} \]

Understanding Moment Generating Functions (mgf)

Recall that the Moment Generating Function (mgf) of a random variable \( X \), denoted as \( M(t) \), is defined as \[ M(t) = E[e^{tX}] \] So we can replace the \( E[e^{tX}] \) in the inequality derived in step 1 with \( M(t) \), yielding \[ P(X \geq a) \leq \frac{M(t)}{e^{ta}} \]

Simplifying the Expression

The final step is to simplify this relationship. We can move terms around to arrive at the desired formula \[ P(X \geq a) \leq e^{-at} M(t), \quad 0 < t < h \] This result shows that the probability of \( X \) being greater than or equal to \( a \) is bounded by the mgf \( M(t) \) scaled by an exponential term involving \( t \) and \( a \).

Apply Similar Steps for P(X ≤ a)

For the second part of the question where \( -h < t < 0 \), we use a similar argument with \( u(x) = e^{tx} \) and \( c = e^{ta} \). Using the same methodology, we can show that \[ P(X \le a) \leq e^{-at} M(t), \quad -h < t < 0 \]

Key concepts

Probability Bounds

When discussing probability bounds, we often want to find the limits on the probabilities of certain outcomes for a random variable. These bounds serve as a way to understand what is the maximum or minimum probability that an event can occur.

In the exercise, we aim to establish probability bounds for events involving a random variable, specifically those indicating that the random variable is greater than or equal to a value or less than or equal to a value.

Given a random variable \(X\) with a moment-generating function (mgf), we derive these bounds for two events:

• \(P(X \geq a)\): the probability that \(X\) is at least \(a\)
• \(P(X \leq a)\): the probability that \(X\) is at most \(a\)

By using a suitable theorem, we discover that both probabilities can be constrained by expressions involving the mgf. This powerful technique allows us to determine rigorous probability bounds without knowing the specific distribution of \(X\).

Understanding these bounds helps you to evaluate risks and make predictions regarding a random variable, without needing detailed information about its distribution.

Theorem Application

Applying theorems correctly is crucial in probability and statistics. In our exercise, theorem application plays a key role in reflecting how advanced mathematical concepts can give us insights into probabilities.

We use Theorem 1.10.2, which connects the expected value of functions of a random variable with probability bounds. The theorem states a probability inequality given a function \(u(x)\) and a constant \(c\):

• \(P(X > a) \leq \frac{E[u(X)]}{u(c)}\)

To apply this theorem in our exercise, we chose an exponential function \(u(x) = e^{tx}\) and a constant \(c = e^{ta}\).

By substituting these into the theorem's format, we link mgf to the expression derived, transforming it into the desired probability bounds we sought.

This example showcases how a powerful theorem can simplify complex probability expressions by substituting appropriate functions and constants matching the question's requirement.

Moment Generating Function in Inequality Proofs

Moment generating functions (mgf) are versatile tools in probability. They offer a way to capture all moments (expected values) of a random variable into a single function. In inequality proofs, mgf can help us derive bounds on probabilities by allowing us to express these probabilities in terms of expectations.

In the given exercise, the mgf of a random variable \(X\), denoted as \(M(t)\), is crucial in simplifying the bounds:

• \(M(t) = E[e^{tX}]\)

Utilizing the mgf, we incorporate it into the steps to arrive at:

• For \(P(X \geq a)\): replace \(E[e^{tX}]\) with \(M(t)\) to get \(P(X \geq a) \leq \frac{M(t)}{e^{ta}}\)
• For \(P(X \leq a)\): again, \(E[e^{tX}]\) is replaced in a similar fashion \(P(X \leq a) \leq \frac{M(t)}{e^{ta}}\)

These relationships are further simplified to obtain the bounds involving exponential terms, revealing how mgf serves in deriving inequalities.

Thus, the moment generating function not only helps in defining the properties of a distribution but also significantly aids in proving inequalities that involve complex expressions in probability.