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Suppose that \(X\) and \(Y\) are \(n - \)dimensional random vectors for which the mean vectors \(E(X)\) and \(E(Y)\) exist . Show that \(E(X + Y) = E(X) + E(Y)\).

Short Answer

Expert verified

By using the property for random variables we proved that \(E(X + Y) = E(X) + E(Y)\).

Step by step solution

01

Define linear statistical models

A linear model describes the correlation between the dependent and independent variables as a straight line.

\(y = {a_0} + {a_1}{x_1} + {a_2}{x_2} + \tilde A,\hat A1/4 + {a_n}{x_n}\)

Models using only one predictor are simple linear regression models. Multiple predictors are used in multiple linear regression models. For many response variables, multiple regression analysis models are used.

02

Find the proof of the given statement

\(X\)and \(Y\) are \(n - \)dimensional random vectors.

Let,

\(\begin{array}{*{20}{c}}{X = \left[ {\begin{array}{*{20}{c}}{{X_1}}\\ \cdots \\{{X_n}}\end{array}} \right]}\\{X = \left[ {\begin{array}{*{20}{c}}{{Y_1}}\\ \cdots \\{{Y_n}}\end{array}} \right]}\end{array}\)

For random variables we know that \(E(X + Y) = E(X) + E(Y)\).

\(\begin{array}{*{20}{c}}{E(X + Y) = E\left( {\left[ {\begin{array}{*{20}{c}}{{X_1} + {Y_1}}\\ \cdots \\{{X_n} + {Y_n}}\end{array}} \right]} \right)}\\{ = \left[ {\begin{array}{*{20}{c}}{E\left( {{X_1} + {Y_1}} \right)}\\ \cdots \\{E\left( {{X_n} + {Y_n}} \right)}\end{array}} \right]}\end{array}\)

\(\begin{array}{*{20}{c}}{ = \left[ {\begin{array}{*{20}{c}}{E\left( {{X_1}} \right) + E\left( {{Y_1}} \right)}\\ \cdots \\{E\left( {{X_n}} \right) + E\left( {{Y_n}} \right)}\end{array}} \right]}\\{ = \left[ {\begin{array}{*{20}{c}}{E\left( {{X_1}} \right)}\\ \cdots \\{E\left( {{X_n}} \right)}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{E\left( {{Y_1}} \right)}\\ \cdots \\{E\left( {{Y_n}} \right)}\end{array}} \right]}\end{array}\)

\(\begin{array}{*{20}{c}}{ = E\left( {\left[ {\begin{array}{*{20}{c}}{{X_1}}\\ \cdots \\{{X_n}}\end{array}} \right]} \right) + E\left( {\left[ {\begin{array}{*{20}{c}}{{Y_1}}\\ \cdots \\{{Y_n}}\end{array}} \right]} \right)}\\{ = E(X) + E(Y)}\end{array}\)

as was shown.

Hence the given statement is proved.

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Most popular questions from this chapter

Consider a state lottery game in which each winning combination and each ticket consists of one set of k numbers chosen from the numbers 1 to n without replacement. We shall compute the probability that the winning combination contains at least one pair of consecutive numbers.

a. Prove that if\({\bf{n < 2k - 1}}\), then every winning combination has at least one pair of consecutive numbers. For the rest of the problem, assume that\({\bf{n}} \le {\bf{2k - 1}}\).

b. Let\({{\bf{i}}_{\bf{1}}}{\bf{ < }}...{\bf{ < }}{{\bf{i}}_{\bf{k}}}\)be an arbitrary possible winning combination arranged in order from smallest to largest. For\({\bf{s = 1,}}...{\bf{,k}}\), let\({{\bf{j}}_{\bf{s}}}{\bf{ = }}{{\bf{i}}_{\bf{s}}}{\bf{ - }}\left( {{\bf{s - 1}}} \right)\). That is,

\(\begin{array}{c}{{\bf{j}}_{\bf{1}}}{\bf{ = }}{{\bf{i}}_{\bf{1}}}\\{{\bf{j}}_{\bf{2}}}{\bf{ = }}{{\bf{i}}_{\bf{2}}}{\bf{ - 1}}\\{\bf{.}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{j}}_{\bf{k}}}{\bf{ = }}{{\bf{i}}_{\bf{k}}}{\bf{ - }}\left( {{\bf{k - 1}}} \right)\end{array}\)

Prove that\(\left( {{{\bf{i}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{i}}_{\bf{k}}}} \right)\)contains at least one pair of consecutive numbers if and only if\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)contains repeated numbers.

c. Prove that\({\bf{1}} \le {{\bf{j}}_{\bf{1}}} \le ... \le {{\bf{j}}_{\bf{k}}} \le {\bf{n - k + 1}}\)and that the number of\(\left( {{{\bf{j}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{j}}_{\bf{k}}}} \right)\)sets with no repeats is\(\left( {\begin{array}{*{20}{c}}{{\bf{n - k + 1}}}\\{\bf{k}}\end{array}} \right)\)

d. Find the probability that there is no pair of consecutive numbers in the winning combination.

e. Find the probability of at least one pair of consecutive numbers in the winning combination

Prove that Following no. is an Integer

\(\frac{{4155 \times 4156 \times ....4250 \times 4251}}{{2 \times 3 \times .... \times 96 \times 97}}\)

A deck of 52 cards contains four aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all four aces will be received by the same

player?

Let \({A_{1,}}{A_{2...}}\) be an infinite sequence of events such that \(A \subset {A_2} \subset ...\) Prove that \({\rm P}\left( {\bigcup\limits_{i = 1}^\infty {{{\rm A}_i}} } \right) = \mathop {lim}\limits_{n \to \infty } {\rm P}\left( {{{\rm A}_n}} \right)\)

The solution to Exercise 1 of Sec. 3.9 is the pdf. of \({{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}\)in Example 6.1.2. Find the pdf. of\({{\bf{\bar X}}_{\bf{2}}}{\bf{ = }}\frac{{{{\bf{X}}_{\bf{1}}}{\bf{ + }}{{\bf{X}}_{\bf{2}}}}}{{\bf{2}}}\). Compare the probabilities that \({X_2}\) and \({X_1}\)are close to 0.5. In particular, compute \({\bf{P}}\left( {\left| {{{\bf{X}}_{\bf{2}}}{\bf{ - 0}}{\bf{.5}}} \right|{\bf{ < 0}}{\bf{.1}}} \right)\)and\({\bf{P}}\left( {\left| {{{\bf{X}}_{\bf{1}}}{\bf{ - 0}}{\bf{.5}}} \right|{\bf{ < 0}}{\bf{.1}}} \right)\). What feature of the pdf \({X_2}\)makes it clear that the distribution is more concentrated near the mean?

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