Question

Calculate the moment generating function of the uniform distribution on \((0,1)\). Obtain \(E[X]\) and \(\operatorname{Var}[X]\) by differentiating.

Short answer

The moment generating function (MGF) of the uniform distribution on \((0,1)\) is \(M_X(t) = \frac{1}{t}(e^t - 1)\). By differentiating, we obtain \(E[X] = \frac{1}{2}\) and \(Var[X] = \frac{1}{12}\).

Step-by-step solution

Find the Probability Density Function (PDF) of X

For the uniform distribution on the interval \((0,1)\), the PDF is as follows: \[f_X(x) = \begin{cases} 1 & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases} \]

Define and Calculate the Moment Generating Function (MGF) of X

The MGF of a random variable \(X\) with PDF \(f_X(x)\) is denoted as \(M_X(t)\) and defined as: \[M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx} f_X(x) dx\] Now, we will calculate the MGF of X using the PDF we derived in the previous step: \[M_X(t) = \int_{0}^{1} e^{tx} dx\] To integrate, we will use the formula for the integral of an exponential function: \(\int e^{ax} dx = \frac{1}{a}e^{ax} + C\). Thus, we have: \[M_X(t) = \frac{1}{t} (e^{t} - 1)\]

Find E[X] by Differentiating MGF of X

Now, differentiate \(M_X(t)\) with respect to \(t\) and evaluate it at \(t = 0\) to find the expected value of \(X\). \[E[X] = \left.\frac{dM_X(t)}{dt}\right|_{t=0}\] Differentiating, we have: \[\frac{dM_X(t)}{dt} = \frac{e^{t}(t - 1) + 1}{t^2}\] Now, evaluate at \(t = 0\): \[E[X] = \left. \frac{e^{t}(t - 1) + 1}{t^2}\right|_{t=0} = \frac{1}{2}\] Thus, the expected value of \(X\) is \(\frac{1}{2}\).

Find Var[X] by Differentiating MGF of X

Now, differentiate the MGF of X once more w.r.t. \(t\) to find \(E[X^2]\) and then use the formula \(Var[X] = E[X^2] - (E[X])^2\) to calculate the variance of \(X\). First, find \(E[X^2]\) by differentiating the MGF of X: \[E[X^2] = \left.\frac{d^2M_X(t)}{dt^2}\right|_{t=0}\] Differentiating, we have: \[\frac{d^2M_X(t)}{dt^2} = -\frac{e^{t}(t^2-2t+2) - 2}{t^3}\] Now, evaluate at \(t = 0\): \[E[X^2] = \left. -\frac{e^{t}(t^2-2t+2) - 2}{t^3}\right|_{t=0} = \frac{1}{3}\] Now, calculate the variance of \(X\) using the formula: \[Var[X] = E[X^2] - (E[X])^2 = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{12}\] Thus, the variance of \(X\) is \(\frac{1}{12}\).

Key concepts

Probability Density Function

The Probability Density Function (PDF) serves as a cornerstone for understanding the distribution of continuous random variables. The PDF is a function that describes the likelihood of a random variable falling within particular ranges of values. In other words, it gives the density of the probability mass at each value of the random variable.

For the uniform distribution on the interval \(0,1\), as highlighted in the exercise, the PDF is elegantly simple. It's defined as being 1 for all values of \(x\) between 0 and 1 and 0 elsewhere. This implies that every point in the interval \(0,1\) is equally likely; there's a uniform distribution of probability across the entire range.

Visualizing the PDF

Imagine a straight horizontal line at height 1 from \(x=0\) to \(x=1\), and that's the graphical representation of the PDF for a uniform distribution. There are no peaks or valleys, illustrating that each outcome within \(0,1\) has the same chance of occurring. This constant function makes calculations involving the PDF, such as finding the expected value or variance, straightforward relative to more complex distributions.

Expected Value

The expected value of a random variable represents the long-run average value of repetitions of the same experiment it describes. For a discrete random variable, the expected value is calculated by summing the products of all possible values of the variable and their respective probabilities. In the context of continuous random variables, however, like the one described by the uniform distribution in our exercise, expected value is found by integrating the product of the variable's value and its PDF over the range of all possible values.

In our case, the expected value \(E[X]\) of a uniform distribution on the interval \(0,1\) is \(\frac{1}{2}\). This result aligns perfectly with our intuitive understanding of symmetry in a uniform distribution since the middle of the range \(0,1\) is exactly \(\frac{1}{2}\).

Understanding the Expected Value

Think of the expected value as the balance point of the PDF's graph. For a uniform distribution on \(0,1\), if you were to physically balance the line segment representing the PDF on a pivot, it would balance exactly at \(\frac{1}{2}\), illustrating the concept of the expected value visually.

Variance of Random Variable

Variance is a numerical value that describes the spread of a random variable's outcomes around the mean (expected value). For a random variable \(X\), the variance is denoted as \(Var[X]\) and provides insight into the degree of dispersion or concentration of the distribution. A smaller variance indicates that the random variable's outcomes are bunched closer to the mean, while a larger variance suggests a wider spread.

As detailed in the solution, to calculate the variance of a uniform distribution on the interval \(0,1\), we use the moment generating function (MGF) to first compute \(E[X^2]\), the expected value of the squared variable. Then, we apply the formula \(Var[X] = E[X^2] - (E[X])^2\) to find the variance. For our uniform distribution, the variance is \(\frac{1}{12}\).

The Significance of Variance

Variance tells us that, in the context of a uniform distribution between 0 and 1, the outcomes are spread evenly across the range, but they're not too far spread out. The variance of \(\frac{1}{12}\) suggests a moderate level of dispersion that is characteristic of the uniform distribution's flat, equal probabilities across the interval. It's a clear measure of variability that complements the intuitive understanding a fixed width of 1 between the maximum and minimum values of \(X\).