Chapter 7: Q.7.42 (page 362)

It follows from Proposition $6.1$ and the fact that the best linear predictor of $Y$ with respect to $X$ is $μ y + ρ σ y / σ x (X - μ x)$ that if $E [Y | X] = a + b X$ then $a = μ y - ρ σ y / σ x μ x$ $b = ρ σ y σ x$ (Why?) Verify this directly

Short Answer

Expert verified

Minimize the expected squared error.

Step by step solution

Given Information

$a = μ_{y} - ρ \frac{σ_{y}}{σ_{x}} μ_{x}$ , $b = ρ \frac{σ_{y}}{σ_{x}}$

Explanation

Let's find constants $a$ and $b$ such that minimizes the expected squared error

E ((Y - a - b X)^{2})

If we differentiate it partially respective to $a$ , we end up with condition $a + b E (X) = E (Y)$

and if we differentiate it partially respective to $b$ , we end up with condition

a E (X) + b E (X^{2}) = E (X Y)

Explanation

The system of equations,

$\{\begin{cases} a & + b E (X) & = E (Y) \\ a E (X) & + b E (X^{2}) & = E (X Y) \end{cases}$

and if we solve it, we get

b = \frac{Cov (X, Y)}{Var (X)}

$= ρ \cdot \frac{σ_{y}}{σ_{x}}$

and

a = μ_{y} - μ_{x} \cdot b = μ_{y} - μ_{x} \cdot ρ \cdot \frac{σ_{y}}{σ_{x}}

Final Answer

Minimize the expected squared error. Hence, we have proved the claimed

a = μ_{y} - μ_{x} \cdot b = μ_{y} - μ_{x} \cdot ρ \cdot \frac{σ_{y}}{σ_{x}}

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It follows from Proposition $6.1$ and the fact that the best linear predictor of $Y$ with respect to $X$ is $μ y + ρ σ y / σ x (X - μ x)$ that if $E [Y | X] = a + b X$ then $a = μ y - ρ σ y / σ x μ x$ $b = ρ σ y σ x$ (Why?) Verify this directly

Short Answer

Step by step solution

Given Information

Explanation

Explanation

Final Answer

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It follows from Proposition 6.1 and the fact that the best linear predictor of Y with respect to X is μy+ρσy/σx(X−μx) that if E[Y|X]=a+bX then a=μy−ρσy/σxμx b=ρσyσx (Why?) Verify this directly

Short Answer

Step by step solution

Given Information

Explanation

Explanation

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Study anywhere. Anytime. Across all devices.

It follows from Proposition $6.1$ and the fact that the best linear predictor of $Y$ with respect to $X$ is $μ y + ρ σ y / σ x (X - μ x)$ that if $E [Y | X] = a + b X$ then $a = μ y - ρ σ y / σ x μ x$ $b = ρ σ y σ x$ (Why?) Verify this directly