Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 15

The definition of the variance of a random variable can be used to show a number of additional results. a. Show that \(\operatorname{Var}(x)=E\left(x^{2}\right)-[E(x)]^{2}\) b. Use Markov's inequality (Problem \(2.14 \mathrm{d}\) ) to show that if \(x\) can take on only non-negative values, \\[ P\left[\left(x-\mu_{x}\right) \geq k\right] \leq \frac{\sigma_{x}^{2}}{k^{2}} \\] This result shows that there are limits on how often a random variable can be far from its expected value. If \(k=h \sigma\) this result also says that \\[ P\left[\left(x-\mu_{x}\right) \geq h \sigma\right] \leq \frac{1}{h^{2}} \\]. Therefore, for example, the probability that a random variable can be more than two standard deviations from its expected value is always less than \(0.25 .\) The theoretical result is called Chebyshev's inequality. c. Equation 2.197 showed that if two (or more) random variables are independent, the variance of their sum is equal to the sum of their variances. Use this result to show that the sum of \(n\) independent random variables, each of which has expected value \(\mu\) and variance \(\sigma^{2},\) has expected value \(m \mu\) and variance \(n \sigma^{2}\). Show also that the average of these \(n\) random variables (which is also a random variable) will have expected value \(\mu\) and variance \(\sigma^{2} / n\). This is sometimes called the law of large numbers-that is, the variance of an average shrinks down as more independent variables are included. d. Use the result from part (c) to show that if \(x_{1}\) and \(x_{2}\) are independent random variables each with the same expected value and variance, the variance of a weighted average of the two \(X=k x_{1}+(1-k) x_{2}, 0 \leq k \leq 1\) is minimized when \(k=0.5\) How much is the variance of this sum reduced by setting \(k\) properly relative to other possible values of \(k\) ? e. How would the result from part (d) change if the two variables had unequal variances?

Short Answer

Expert verified
Answer: The variance of the weighted average of two independent random variables is minimized when the weighting factor, k, is equal to 0.5.

Step by step solution

01

Recall the definition of variance

The variance of a random variable x is given by: \(Var(x) = E\left[(x - E(x))^2\right]\)
02

Expand the expression inside the expectation

Expand the expression \((x - E(x))^2\): \((x - E(x))^2 = x^2 - 2xE(x) + [E(x)]^2\)
03

Use the linearity of expectation

The linearity of expectation states that for any random variables X and Y, E(X+Y) = E(X) + E(Y). Apply this to our expanded expression: \(Var(x) = E\left[x^2 - 2xE(x) + [E(x)]^2\right] = E(x^2) - 2E(x)E(x) + E([E(x)]^2)\)
04

Simplify the expression

Simplify the expression to get the desired result: \(Var(x) = E(x^2) - 2[E(x)]^2 + [E(x)]^2 = E(x^2) - [E(x)]^2\) b. Show that \(P\left[\left(x-\mu_{x}\right) \geq k\right] \leq \frac{\sigma_{x}^{2}}{k^{2}}\)
05

Recall Markov's inequality

Markov's inequality states that for any non-negative random variable X and any positive value k, \(P\left[X \geq k\right] \leq \frac{E(X)}{k}\).
06

Apply Markov's inequality to given expression

Let Y = \((x - E(x))^2\). Then Markov's inequality becomes: \(P\left[(x - \mu_x)^2 \geq k^2\right] \leq \frac{E\left[(x - \mu_x)^2\right]}{k^2}\)
07

Use the definition of variance

We know that the variance of a random variable is \(Var(x) = E\left[(x - \mu_x)^2\right] = \sigma_x^2\). Substitute this into the inequality: \(P\left[(x - \mu_x)^2 \geq k^2\right] \leq \frac{\sigma_x^2}{k^2}\)
08

Change the expression inside the probability

We need to change the expression inside the probability to match the given expression: \(P\left[\left(x-\mu_{x}\right) \geq k\right] \leq \frac{\sigma_{x}^{2}}{k^{2}}\) c. Find the expected value and variance of the sum and average of \(n\) independent random variables.
09

Find the expected value of the sum of random variables

Let \(X_1, X_2, \dots, X_n\) be the n independent random variables, each with the expected value m and variance \(\sigma^2\). The expected value of their sum \(S = X_1 + X_2 + \dots + X_n\) is: \(E(S) = E(X_1 + X_2 + \dots + X_n) = E(X_1) + E(X_2) + \dots + E(X_n) = nm\)
10

Find the variance of the sum of random variables

The variance of their sum is: \(Var(S) = Var(X_1 + X_2 + \dots + X_n) = Var(X_1) + Var(X_2) + \dots + Var(X_n) = n\sigma^2\)
11

Find the expected value and variance of the average of random variables

The average of the random variables \(A = \frac{1}{n}(X_1 + X_2 + \dots + X_n)\). The expected value and variance of the average are: \(E(A) = E\left(\frac{1}{n}S\right) = \frac{1}{n}E(S) = m\) \(Var(A) = Var\left(\frac{1}{n}S\right) = \frac{1}{n^2}Var(S) = \frac{\sigma^2}{n}\) d. Minimize the variance of the weighted average of two independent random variables.
12

Write the equation for the weighted average

Let \(X_1\) and \(X_2\) be two independent random variables, both with the same expected value and variance. The weighted average X is given by \(X = kX_1 + (1-k)X_2\), where \(0 \leq k \leq 1\).
13

Calculate the variance of the weighted average

The variance of the weighted average is: \(Var(X) = Var(kX_1 + (1-k)X_2) = k^2Var(X_1) + (1-k)^2Var(X_2) = k^2\sigma^2 + (1-k)^2\sigma^2\)
14

Differentiate the variance of the weighted average with respect to k

To minimize the variance of the weighted average, we need to find the value of k that minimizes the expression. Differentiate \(Var(X)\) with respect to k: \(\frac{dVar(X)}{dk} = 2k\sigma^2 - 2(1-k)\sigma^2 = 4k\sigma^2 - 2\sigma^2\)
15

Find the value of k that minimizes the variance of the weighted average

Set the derivative to zero and solve for k: \(4k\sigma^2 - 2\sigma^2 = 0\) \(k = \frac{1}{2}\) So the variance of the weighted average is minimized when \(k = 0.5\). e. If the two variables had unequal variances, the optimization of the value of k that minimizes the variance of the weighted average would be different and needs to be found by differentiating the new expression for the weighted average with respect to k.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variance of a Random Variable
Understanding the variance of a random variable is crucial as it measures the spread of a set of values. In simpler terms, it tells us how much the numbers in a data set differ from the average (mean) value of the set. Let's break this down with an example: Imagine you have test scores from a class. The variance would help you understand how consistent the students' performances were. If the variance is high, students scored very differently from each other, while a low variance indicates that most students scored similarly to the average.

Using the formula \( Var(x) = E(x^2) - [E(x)]^2 \), we're essentially looking at the expected value of the squared difference between each data point and the mean. This squared difference emphasizes larger deviations from the mean, showing us how 'spread out' the scores are. Always keep in mind, the higher the variance, the more volatile or risky a set of scores or values is considered to be.
Markov's Inequality
The Markov's inequality plays a crucial role in probability and statistics, basically telling us that rare events do not happen frequently. If we take an example of people's heights in a population, Markov's inequality would give us an upper bound for the probability that a person is extremely tall. For instance, if the average height is 1.7 meters, the inequality can help us understand the likelihood that someone is over 3 meters tall, which would be exceedingly rare. The inequality \( P[X \geq k] \leq \( E(X) / k \) \) simply says that the probability of our random variable being at least 'k' is less than or equal to the expected value of the variable divided by 'k'. This concept is very handy when dealing with non-negative random variables to put a ceiling on probabilities.
Chebyshev's Inequality
Expanding upon the theme of understanding variability, Chebyshev's inequality is a bit like the statistical safety net, ensuring that even in a diverse set of data, most values are close to the mean. To visualize this, picture a lake with buoys spread across — no matter how wide the lake is, Chebyshev's inequality guarantees that a large percentage of these buoys will be within a certain distance from the center.

The formula \( P[|(x - \mu_x)| \geq k] \leq \( \sigma_x^2 / k^2 \) \) provides the comfort that, regardless of the distribution shape, only a small fraction of values will be more than 'k' standard deviations away from the mean. This is incredibly useful because it works for a wide range of distributions, including those that are not perfectly bell-shaped.
Law of Large Numbers
The law of large numbers is a bit like making a quilt — the more patches (or data points) you add, the smoother and more even it looks. In statistics, it tells us that as we increase our sample size, the average of the sample gets closer and closer to the expected value. Imagine flipping a coin; the more you flip, the average number of heads and tails will get closer to an equal split, even if you start with a streak of all heads or all tails.

When we sum up a large number of independent random variables, all with the same average and variability, we can expect the sum to have a predictable average, but the variance of the averages decreases with more variables. This is essential for predicting outcomes in various situations, such as manufacturing, survey results, and even gambling. As the number of observations grows, the cumulative effect becomes more stable and reliable.
Weighted Average
Consider the weighted average as a teacher grading a project with various parts having different importance. Some sections might count more towards the final grade than others; this is the essence of weighting. In statistics, a weighted average gives us a way to combine values where some points carry more significance. For instance, if you're measuring school performance, test scores (weighted heavily) might influence the overall score more than homework (weighted lightly).

Mathematically, when we combine independent random variables with different 'weights', the outcome depends on those weights. It's like a balance scale — put too much weight on one side (or one variable), and it skews the results. By optimizing these weights (often equally, as seen in the case with variables with the same variance), we minimize the variability of our weighted average, leading to a more stable and accurate representation of the underlying data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Another function we will encounter often in this book is the power function: \\[ y=x^{\delta} \\] the form \(y=x^{8} / 8\) to ensure that the derivatives have the proper sign). a. Show that this function is concave (and therefore also, by the result of Problem \(2.9,\) quasi-concave). Notice that the \(\delta=1\) is a special case and that the function is "strictly" concave only for \(\delta<1\) b. Show that the multivariate form of the power function \\[ y=f\left(x_{1}, x_{2}\right)=\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8} \\] is also concave (and quasi-concave). Explain why, in this case, the fact that \(f_{12}=f_{21}=0\) makes the determination of concavity especially simple. c. One way to incorporate "scale" effects into the function described in part (b) is to use the monotonic transformation \\[ g\left(x_{1}, x_{2}\right)=y^{y}=\left[\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}\right]^{y} \\] where \(\gamma\) is a positive constant. Does this transformation preserve the concavity of the function? Is \(g\) quasi-concave?

Suppose \(U(x, y)=4 x^{2}+3 y^{2}\) a. Calculate \(\partial U / \partial x, \partial U / \partial y\) b. Evaluate these partial derivatives at \(x=1, y=2\) c. Write the total differential for \(U\) d. Calculate \(d y / d x\) for \(d U=0\) -that is, what is the implied trade-off between \(x\) and \(y\) holding \(U\) constant? e. Show \(U=16\) when \(x=1, y=2\) f. In what ratio must \(x\) and \(y\) change to hold \(U\) constant at 16 for movements away from \(x=1, y=2 ?\) g. More generally, what is the shape of the \(U=16\) contour line for this function? What is the slope of that line?

involving the function and its derivatives. Here we look at some applications of the theorem for functions of one and two variables. a. Any continuous and differentiable function of a single variable, \(f(x),\) can be approximated near the point \(a\) by the formula \\[ f(x)=f(a)+f^{\prime}(a)(x-a)+0.5 f^{\prime \prime}(a)(x-a)^{2}+\text { terms in } f^{\prime \prime \prime}, f^{\prime \prime \prime \prime}, \ldots \\] Using only the first three of these terms results in a quadratic Taylor approximation. Use this approximation together with the definition of concavity given in Equation 2.85 to show that any concave function must lie on or below the tangent to the function at point \(a\) b. The quadratic Taylor approximation for any function of two variables, \(f(x, y),\) near the point \((a, b)\) is given by \\[ \begin{aligned} f(x, y)=& f(a, b)+f_{1}(a, b)(x-a)+f_{2}(a, b)(y-b) \\ &+0.5\left[f_{11}(a, b)(x-a)^{2}+2 f_{12}(a, b)(x-a)(y-b)+f_{22}(y-b)^{2}\right] \end{aligned} \\] Use this approximation to show that any concave function (as defined by Equation 2.98 ) must lie on or below its tangent plane at \((a, b)\).

A simple way to model the construction of an oil tanker is to start with a large rectangular sheet of steel that is \(x\) feet wide and \(3 x\) feet long. Now cut a smaller square that is \(t\) feet on a side out of each corner of the larger sheet and fold up and weld the sides of the steel sheet to make a traylike structure with no top. a. Show that the volume of oil that can be held by this tray is given by \\[ V=t(x-2 t)(3 x-2 t)=3 t x^{2}-8 t^{2} x+4 t^{3} \\] b. How should \(t\) be chosen to maximize \(V\) for any given value of \(x ?\) c. Is there a value of \(x\) that maximizes the volume of oil that can be carried? d. Suppose that a shipbuilder is constrained to use only 1,000,000 square feet of steel sheet to construct an oil tanker. This constraint can be represented by the equation \(3 x^{2}-4 t^{2}=1,000,000\) (because the builder can return the cut-out squares for credit). How does the solution to this constrained maximum problem compare with the solutions described in parts and (c)?

The height of a ball that is thrown straight up with a certain force is a function of the time ( \(t\) ) from which it is released given by \(f(t)=-0.5 g t^{2}+40 t\) (where \(g\) is a constant determined by gravity). a. How does the value of \(t\) at which the height of the ball is at a maximum depend on the parameter \(g\) ? b. Use your answer to part (a) to describe how maximum height changes as the parameter \(g\) changes. c. Use the envelope theorem to answer part (b) directly. d. On the Earth \(g=32\), but this value varies somewhat around the globe. If two locations had gravitational constants that differed by \(0.1,\) what would be the difference in the maximum height of a ball tossed in the two places?
See all solutions

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free