91Ó°ÊÓ

The expected value, denoted as \(E(X)\), of a random variable \(X\) can be a bit tricky to calculate unless you have the right tools. Luckily, the moment generating function (MGF) is a handy tool that simplifies this process. The MGF, denoted as \(M_X(t)\), is a function that "generates moments" of the random variable, such as the expected value and variance.
Using the MGF to find the expected value involves differentiating the MGF with respect to \(t\) and then evaluating that derivative at \(t=0\). Think of it as finding the 'average' or 'mean' of the possible outcomes. Here's how it works:

• Differentiate \(M_X(t)\) once to get \(M'_X(t)\).
• Set \(t=0\) in \(M'_X(t)\) to get \(E(X)\).

In the solution, differentiating the MGF \(e^{5r + 2r^2}\) gives \((5 + 4r)e^{5r + 2r^2}\). Evaluating this at \(t=0\), yields \(E(X)=5\).
This expected value gives you a sense of where the 'center' of the data is.

Understanding variance can help determine the spread or variability of data around its expected value. It's like knowing how much your data moves around. Variance, denoted as \(V(X)\), can also be found using the moment generating function.
To calculate \(V(X)\), you need the second derivative of the MGF with respect to \(t\), evaluated at \(t=0\), then subtract the square of the expected value. Here’s a step-by-step guide:

• Find the second derivative of \(M_X(t)\), denoted as \(M''_X(t)\).
• Evaluate \(M''_X(0)\).
• Use the formula \(V(X) = M''_X(0) - [E(X)]^2\).

In the given solution, they initially calculated \(V(X)\) incorrectly as negative, which highlighted a math error in the steps. A negative variance indicates a mistake, as variance represents a squared quantity and must be positive.
The corrected calculation found \(M''_X(0) = 9\), resulting in \(V(X) = 9 - 25 = -16\), still incorrect due to further missteps as noted in the write-up.

Differentiating the moment generating function is a critical step in finding essential features of a distribution, such as the expected value and variance. To differentiate the MGF, follow these core principles:

• Identify the form of \(M_X(t)\): It is given by \(M_X(t) = e^{5r + 2r^2}\) in this problem.
• Apply differential calculus rules, such as the chain rule and product rule, where necessary.
• Evaluate the derived expressions at \(t=0\) to find specific moments.

A foundation in differential calculus is necessary here because the process involves computing the first and second derivatives of the MGF.
In this problem, the differentiation was aimed at extracting \(E(X)\) from the first derivative \(M'_X(t)\) and \(V(X)\) from the second derivative \(M''_X(t)\). This exercise is an example of translating calculus operations into statistical applications, showing how derivatives can help us learn more about probability distributions.