Finding E(X³) from the Moment Generating Function

The question asks us to find the third moment about the origin, denoted as \(E(X^3)\), for a continuous random variable X. We are given its moment generating function (MGF), \(M_X(t)\). The moment generating function is a powerful tool in probability theory that allows us to find moments of a random variable.

Understanding the Moment Generating Function (MGF)

The moment generating function (MGF) of a random variable X is defined as \(M_X(t) = E(e^{tX})\) for all values of t for which the expectation exists. A key property of the MGF is that the moments about the origin can be obtained by taking derivatives of the MGF with respect to t and evaluating them at \(t=0\). Specifically, the k-th moment \(E(X^k)\) is given by:

$$ E(X^k) = \frac{d^k}{dt^k} M_X(t) \Bigg|_{t=0} $$

In this problem, we need to find \(E(X^3)\), which means we need to find the third derivative of the given MGF with respect to t and then set \(t=0\).

Given Moment Generating Function

The given MGF for the continuous random variable X is:

$$ M_X(t) = \frac{\lambda}{\lambda-t}, \quad t < \lambda $$

We can rewrite this function using negative exponents to make differentiation easier:

$$ M_X(t) = \lambda (\lambda-t)^{-1} $$

Step-by-Step Calculation of E(X³)

We need to find the first, second, and third derivatives of \(M_X(t)\) with respect to t.

First Derivative (\(k=1\)):

$$ \frac{d}{dt} M_X(t) = \frac{d}{dt} [\lambda (\lambda-t)^{-1}] $$

Using the chain rule \(\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)\), where \(f(u) = \lambda u^{-1}\) and \(u = g(t) = \lambda-t\). So \(f'(u) = -\lambda u^{-2}\) and \(g'(t) = -1\).

$$ \frac{d}{dt} M_X(t) = (-\lambda (\lambda-t)^{-2}) \cdot (-1) = \lambda (\lambda-t)^{-2} $$

Evaluating at \(t=0\) gives \(E(X) = \lambda (\lambda-0)^{-2} = \lambda/\lambda^2 = 1/\lambda\).

Second Derivative (\(k=2\)):

Now, differentiate the first derivative:

$$ \frac{d^2}{dt^2} M_X(t) = \frac{d}{dt} [\lambda (\lambda-t)^{-2}] $$

Using the chain rule again, where \(f(u) = \lambda u^{-2}\) and \(u = g(t) = \lambda-t\). So \(f'(u) = -2\lambda u^{-3}\) and \(g'(t) = -1\).

$$ \frac{d^2}{dt^2} M_X(t) = (-2\lambda (\lambda-t)^{-3}) \cdot (-1) = 2\lambda (\lambda-t)^{-3} $$

Evaluating at \(t=0\) gives \(E(X^2) = 2\lambda (\lambda-0)^{-3} = 2\lambda/\lambda^3 = 2/\lambda^2\).

Third Derivative (\(k=3\)):

Finally, differentiate the second derivative to find the third derivative:

$$ \frac{d^3}{dt^3} M_X(t) = \frac{d}{dt} [2\lambda (\lambda-t)^{-3}] $$

Using the chain rule, where \(f(u) = 2\lambda u^{-3}\) and \(u = g(t) = \lambda-t\). So \(f'(u) = -6\lambda u^{-4}\) and \(g'(t) = -1\).

$$ \frac{d^3}{dt^3} M_X(t) = (-6\lambda (\lambda-t)^{-4}) \cdot (-1) = 6\lambda (\lambda-t)^{-4} $$

Evaluate at t=0 to find E(X³):

Now, we evaluate the third derivative at \(t=0\) to find \(E(X^3)\):

$$ E(X^3) = \frac{d^3}{dt^3} M_X(t) \Bigg|_{t=0} = 6\lambda (\lambda-0)^{-4} $$

$$ E(X^3) = 6\lambda (\lambda)^{-4} = \frac{6\lambda}{\lambda^4} = \frac{6}{\lambda^3} $$

Therefore, the expected value of X cubed, \(E(X^3)\), is \(\frac{6}{\lambda^3}\).

Conclusion

Based on the step-by-step derivation using the moment generating function, the value of \(E(X^3)\) for the continuous random variable X with the given MGF is \(\frac{6}{\lambda^3}\).

Revision Table: Moment Generating Function Properties

Additional Information: Connecting MGF to Distributions

The moment generating function \(M_X(t) = \frac{\lambda}{\lambda-t}\) is the MGF of an Exponential distribution with rate parameter \(\lambda\). The Exponential distribution is a continuous probability distribution that describes the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate \(\lambda\). Knowing the MGF allows us to identify the distribution and use its known properties. For an Exponential(\(\lambda\)) distribution, the mean is \(1/\lambda\), the variance is \(1/\lambda^2\), and the k-th moment is \(k! / \lambda^k\). Let's check this formula for \(k=3\): \(E(X^3) = 3! / \lambda^3 = 6 / \lambda^3\), which matches our result obtained from differentiating the MGF. This confirms our calculation and demonstrates the power of the MGF in identifying distributions and their properties.