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Show that if \(\mathbf{E}\left(X^{2}\right)<\infty\), \(\min _{a} \mathbf{E}\left((X-a)^{2}\right)=\operatorname{var}(X)\).

Short Answer

Expert verified
The minimum, \(\operatorname{var}(X)\), occurs when \(a = \mathbf{E}(X)\).

Step by step solution

01

Understand the Problem

The problem asks us to find the value of \(a\) that minimizes the expression \(\mathbf{E}\left((X-a)^{2}\right)\) and show that this minimum value is equal to the variance of \(X\), denoted as \(\operatorname{var}(X)\). Given that the expected value of \(X^2\) is finite, this condition ensures the finiteness required for variance.
02

Rewrite the Expression

Rewrite \(\mathbf{E}((X-a)^2)\) using the definition of variance \(\operatorname{var}(X) = \mathbf{E}(X^2) - (\mathbf{E}(X))^2\). The expression \((X-a)^2\) can be expanded to \(X^2 - 2aX + a^2\).
03

Find the Expectations

Take the expectation of the expanded form: \(\mathbf{E}(X^2 - 2aX + a^2) = \mathbf{E}(X^2) - 2a\mathbf{E}(X) + a^2\). This is the function we need to minimize with respect to \(a\).
04

Differentiate with Respect to a

To find the value of \(a\) that minimizes the function, differentiate \(\mathbf{E}(X^2) - 2a\mathbf{E}(X) + a^2\) with respect to \(a\), which gives the derivative: \(-2\mathbf{E}(X) + 2a\).
05

Solve for the Minimum

Set the derivative equal to zero: \(-2\mathbf{E}(X) + 2a = 0\). Solving for \(a\) gives \(a = \mathbf{E}(X)\).
06

Substitute the Value of a

Substitute \(a = \mathbf{E}(X)\) back into the expression for \(\mathbf{E}((X-a)^2)\). This gives: \(\mathbf{E}((X-\mathbf{E}(X))^2) = \operatorname{var}(X)\) by definition of variance.

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Key Concepts

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

Expectation
In probability theory, the concept of **expectation** is crucial for understanding randomness and statistical predictions. Expected value, often denoted as \( \mathbf{E}(X) \), represents the average or mean value of a random variable over numerous trials. It's like predicting what we expect as an average outcome.

For a discrete random variable, the expectation is calculated by summing the products of each possible value of the variable and their probabilities. Mathematically, it is expressed as:
  • \( \mathbf{E}(X) = \sum x_i p_i \)
where \( x_i \) are the values of the random variable, and \( p_i \) are the probabilities.

For continuous variables, the concept is extended using integrals:
  • \( \mathbf{E}(X) = \int x f(x) dx \)
where \( f(x) \) is the probability density function. Expectation helps in analyzing and forecasting the behavior of a variable, making it indispensable in the derivation and minimization of functions in statistics. In our exercise, setting the expectation of \( X \) helps identify the point where variance, a measure of spread, is minimized.
Minimization
The process of **minimization** seeks to find the smallest value of a function. In the context of our exercise, we strive to minimize \( \mathbf{E}((X-a)^2) \).

Minimization often involves calculus, specifically finding the derivative. This allows us to identify critical points that signify potential minima. By setting the derivative to zero, we find values where the function stops decreasing, theoretically indicating a minimum if confirmed through further analysis, such as sign changes in the second derivative.
  • This process includes mathematically manipulating the function to reveal the most convenient form for differentiation.
  • In our example, this was achieved by rewiring \( \mathbf{E}((X-a)^2) \) as \( \mathbf{E}(X^2) - 2a \mathbf{E}(X) + a^2 \).
  • Then, the derivative concerning \( a \) leads us to \( a = \mathbf{E}(X) \), confirming the expected mean as our optimal point of minimization.
When steps are correctly followed, this ensures the function's minimized point accurately corresponds to key statistical properties like variance.
Function Derivation
**Function derivation** is a method used to find the rate at which a function changes. In mathematics and particularly in optimization problems, derivation aids us in identifying points of interest, such as minima or maxima.

In the context of the exercise, deriving the function \( \mathbf{E}((X-a)^2) \) with respect to \( a \) allows us to find the minimum point that coincides with the variance of \( X \).
  • Derivation involves applying calculus, specifically differentiation techniques.
  • For our expression \( \mathbf{E}(X^2) - 2a \mathbf{E}(X) + a^2 \), the derivative \(-2\mathbf{E}(X) + 2a\) was calculated relative to \( a \).
By setting the derivative to zero, indicative of stationary points, the process isolated \( a = \mathbf{E}(X) \). Thus, the function derivation confirms that placing \( a \) at this mean value minimizes the expectation of the squared differences, a vital part of statistical optimization and variance calculation. Function derivation provides detailed insights into the behavior of a mathematical function by exploring how small changes in inputs affect outputs. This principle underpins a broad array of applications in mathematics, physics, and fields reliant upon data analysis and prediction.

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Most popular questions from this chapter

A random variable is symmetric if for some \(a\) and all \(k, f(a-k)=f(a+k)\). Show that the mean and a median are equal for symmetric random variables. Find a nonsymmetric random variable for which the mean and median are equal.

A belt conveys tomatoes to be packed. Each tomato is defective with probability \(p\), independently of the others. Each is inspected with probability \(r\); inspections are also mutually independent. If a tomato is defective and inspected, it is rejected. (a) Find the probability that the \(n\)th tomato is the \(k\) th defective tomato. (b) Find the probability that the \(n\)th tomato is the \(k\) th rejected tomato. (c) Given that the \((n+1)\) th tomato is the first to be rejected, let \(X\) be the number of its predecessors that were defective. Find \(\mathbf{P}(X=k)\), the probability that \(X\) takes the value \(k\), and \(\mathbf{E}(X)\).

Pascal and Brianchon now play a series of games that may be drawn (i.e., tied) with probability \(r\). Otherwise, Pascal wins with probability \(p\) or Brianchon wins with probability \(q\), where \(p+q+\) \(r=1\). (a) Find the expected duration of the match if they stop when one or other wins two consecutive games. Also, find the probability that Pascal wins. (b) Find the expected duration of the match if they stop when one or other wins two successive games of the games that are won. (That is, draws are counted but ignored.) Find the probability that Pascal wins. If you were Brianchon and \(p>q\), which rules would you rather play by?

Preparatory to a camping trip, you can buy six cans of food, all the same size, two each of meat, vegetables, and fruit. Assuming that cans with the same contents have indistinguishable labels, in how many distinguishable ways can the cans be arranged in a row? On the trip, there is heavy rain and all the labels are washed off. Show that if you open three of the cans at random the chance that you will open one of each type is \(\frac{2}{5}\). If you do not succeed, you continue opening cans until you have one of each type; what is the expected number of open cans?

Mr. Smith must site his widget warehouse in either Acester or Beeley. Initially, he assesses the probability as \(p\) that the demand for widgets is greater in Acester, and as \(1-p\) that it is greater in Beeley. The ideal decision is to site the warehouse in the town with the larger demand. The cost of the wrong decision, because of increased transport costs, may be assumed to be \(£ 1000\) if Acester is the correct choice and \(£ 2000\) if Beeley is the correct choice. Find the expectations of these costs for each of the two possible decisions, and the values of \(p\) for which Acester should be chosen on the basis of minimum expected cost. Mr. Smith could commission a market survey to assess the demand. If Acester has the higher demand, the survey will indicate this with probability \(\frac{3}{4}\) and will indicate Beeley with probability \(\frac{1}{4}\). If Beeley has the higher demand the survey will indicate this with probability \(\frac{2}{3}\) and will indicate Acester with probability \(\frac{1}{3}\). Show the probability that the demand is higher in Acester is \(9 p /(4+5 p)\) if the survey indicates Acester. Find also the expected cost for each of the two possible decisions if the survey indicates Acester. If the survey indicates Acester and \(p<8 / 17\), where should Mr. Smith site the warehouse?

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