Question

If \(0 \leqslant X \leqslant a\), show that (a) \(E\left[X^{2}\right] \leqslant a E[X]\) (b) \(\operatorname{Var}(X) \leqslant E[X](a-E[X])\) (c) \(\operatorname{Var}(X) \leqslant a^{2} / 4\).

Short answer

Given a random variable \(X\) such that \(0 \leqslant X \leqslant a\), we can prove the following inequalities: (a) Using the expected value properties, one can show that \(E[X^2] \leqslant a^2\) and \(E[X] \leqslant a\), which implies that \(E[X^2] \leqslant aE[X]\). (b) One can show that \(\operatorname{Var}(X) \leqslant E[X](a-E[X])\). (c) By finding the maximum value of the function \(f(x) = x(a-x)\) for \(0 \leqslant x \leqslant a\), we obtain that the maximum value is \(f(a/2) = a^2/4\), thus \(\operatorname{Var}(X) \leqslant a^2 / 4\).

Step-by-step solution

(a) Prove that \(E[X^{2}] \leqslant a E[X]\)

To prove \(E[X^2] \leqslant aE[X]\), let's first recall the definition of expected value. The expected value of a function \(g(X)\) of a random variable \(X\) is defined as: \(E[g(X)] = \sum_{i} g(x_i)P(x_i)\) for discrete variables, or \(E[g(X)] = \int_{-\infty}^{\infty} g(x)p(x)dx\) for continuous variables. In this case, we have \(g(X) = X^2\), so we need to compute its expected value: \(E[X^2] = \int_0^a x^2p(x)dx\). Now, since \(0 \leqslant x \leqslant a\), it follows that \(0 \leqslant x^2 \leqslant a^2\). Thus, we can write: \(E[X^2] = \int_0^a x^2p(x)dx \leqslant a^2 \int_0^a p(x)dx\). Since \(p(x)\) is a probability density function, \(\int_0^a p(x)dx\) must equal 1. Therefore: \(E[X^2] \leqslant a^2\). Now let's compute \(E[X]\), and we have \(g(X)=X\): \(E[X] = \int_0^a xp(x)dx\). Since \(0 \leqslant x \leqslant a\), it follows that \(0 \leqslant x \leqslant a\). Thus, we can write: \(E[X] = \int_0^a xp(x)dx \leqslant a \int_0^a p(x)dx\). Again, since \(p(x)\) is a probability density function, \(\int_0^a p(x)dx\) must equal 1. Therefore: \(E[X] \leqslant a\). Now, let's multiply \(E[X]\) by \(a\) and compare it to \(E[X^2]\): \(aE[X] \geqslant a (E[X])^2\). Since \(E[X^2] \leqslant a^2\), and \(aE[X] \geqslant a(E[X])^2\), we conclude that \(E[X^2] \leqslant aE[X]\).

(b) Prove that \(\operatorname{Var}(X) \leqslant E[X](a-E[X])\)

To prove \(\operatorname{Var}(X) \leqslant E[X](a-E[X])\), let's first recall the definition of variance. The variance of a random variable \(X\) is defined as: \(\operatorname{Var}(X) = E[(X - E[X])^2]\). Now, \(E[(X - E[X])^2] = E[X^2 - 2XE[X] + (E[X])^2]\). Using the property of the expected value, we can write: \(\operatorname{Var}(X) = E[X^2] - 2E[X]E[X] + E[(E[X])^2]\). From (a), we know that \(E[X^2] \leqslant aE[X]\), so we have: \(\operatorname{Var}(X) \leqslant aE[X] - 2E[X]E[X] + (E[X])^2 = E[X](a-E[X])\).

(c) Prove that \(\operatorname{Var}(X) \leqslant a^2 / 4\)

In order to prove this inequality, we will use the previous result that \(\operatorname{Var}(X) \leqslant E[X](a-E[X])\). We will now find the maximum possible value for \(E[X](a-E[X])\) under the given constraints. Let \(f(x) = x(a-x)\), where \(0 \leqslant x \leqslant a\). We need to find the maximum value for \(f(x)\). To do this, we can find the critical points by taking the first derivative and setting it to zero: \(f'(x) = a - 2x\) Setting \(f'(x)=0\), we get \(x=a/2\). This critical point is a maximum, as the second derivative \(f''(x)=-2\) is negative. Therefore, the maximum value of \(f(x)\) occurs at \(x = a/2\): \(f(a/2) = \frac{a}{2}\left(a-\frac{a}{2}\right) = \frac{a^2}{4}\). Since \(\operatorname{Var}(X) \leqslant E[X](a-E[X])\), and the maximum value for \(E[X](a-E[X])\) is \(a^2/4\), we conclude that \(\operatorname{Var}(X) \leqslant a^2 / 4\).

Key concepts

Expected Value

The concept of the expected value is fundamental in probability and statistics. It offers a measure of the central tendency of a random variable, essentially representing the average outcome if an experiment were to be repeated a large number of times. For a discrete random variable, the expected value is calculated by summing the products of each outcome with its corresponding probability. In a continuous setting, this summation turns into an integration over the probability density function (PDF).

In the provided exercise, understanding the expected value is critical to solving part (a), where the goal is to show that the expected value of the square of a random variable, denoted as \(E[X^2]\), is less than or equal to the product of a constant \(a\) and the expected value of \(X\), denoted as \(aE[X]\). The stepped solution walks through the process of comparing \(E[X^2]\) to \(aE[X]\) and demonstrates that under the given conditions, the inequality holds true.

Improving comprehension might involve clarifying the conditions under which \(E[X]\) can be calculated and emphasizing the properties of the probability density function involved in the computation of the expected value.

Probability Density Function

A probability density function (PDF) is a statistical expression that defines a probability distribution for a continuous random variable. It is a function \(p(x)\) that maps any given number within the variable's range to the likelihood that the variable's outcome will equal that number. One of the defining properties of a PDF is that the integral over its entire range is equal to one, which mathematically confirms that it encompasses the entire probability distribution.

In the context of the exercise, the PDF is used to calculate the expected values, as seen in the process where \(E[X]\) and \(E[X^2]\) are found by integrating with respect to the PDF over the given range \(0 \leqslant x \leqslant a\). It's crucial for students to grasp that it's this integral, the area under the curve of the PDF, which guarantees that probabilities are normalized and the expected values are meaningful. Including examples of different PDF shapes and how they influence the expected value could be a valuable enhancement to the learning experience.

Variance of a Random Variable

Variance measures how much a set of numbers is spread out or how much they deviate from the expected value (mean). For a random variable, variance gives us an idea of the distribution's dispersion. The variance of a random variable \(X\) is expressed as \(\operatorname{Var}(X)\) and is defined as the expected value of the squared deviation from the mean: \(\operatorname{Var}(X) = E[(X-E[X])^2]\).

In the exercise, the variance is key to understanding parts (b) and (c). Part (b) challenges us to show that \(\operatorname{Var}(X)\) is less than or equal to \(E[X](a-E[X])\), and part (c) involves proving a specific upper limit on \(\operatorname{Var}(X)\). The stepped solution effectively uses the definition of the variance and manipulates the inequality to demonstrate the required results. However, to enhance comprehension, it may be helpful to provide more context on the importance of variance in assessing the reliability of an expected value and to discuss the intuition behind the squared deviations used in its calculation.