Chapter 7: Q.7.17 (page 360)

Suppose that $X_{1}$ and $X_{2}$ are independent random variables having a common mean $μ$ . Suppose also that $Var (X_{1}) = σ_{1}^{2}$ and $Var (X_{2}) = σ_{2}^{2}$ . The value of $μ$ is unknown, and it is proposed that $μ$ be estimated by a weighted average of $X_{1}$ and $X_{2}$ . That is, $λ X_{1} + (1 - λ) X_{2}$ will be used as an estimate of $μ$ for some appropriate value of $λ$ . Which value of $λ$ yields the estimate having the lowest possible variance? Explain why it is desirable to use this value of $λ .$

Short Answer

Expert verified

Reason for it is desirable to use this value of $λ :$

As $Var [λ X_{1} + (1 - λ) X_{2}] = E [{(λ X_{1} + (1 - λ) X_{2})}^{2}]$ then $λ$ is to be small.

Step by step solution

Given Information

Independent Random variables $= X_{1}, X_{2}$

$Var (X_{1}) = σ_{1}^{2}$

$Var (X_{2}) = σ_{2}^{2}$

Value of $μ = ?$

Variance of $λ X_{1} + (1 - λ) X_{2} = ?$

Explanation

Suppose that $X_{1}$ and $X_{2}$ are independent random variables having a common mean $μ$ . Let $λ X_{1} + (1 - λ) X_{2}$ will be used as an estimate of $μ$ for some appropriate value of $λ$ . It is known that $Var (X_{1}) = σ_{1}^{2}$ and $Var (X_{2}) = σ_{2}^{2}$ .

Find the variance of $λ X_{1} + (1 - λ) X_{2}$ .

$Var (λ X_{1} + (1 - λ) X_{2}) = Var (λ X_{1}) + Var [(1 - λ) X_{2}]$

localid="1647409774914" $= λ^{2} Var (X_{1}) + (1 - λ)^{2} Var [X_{2}]$

$= λ^{2} σ_{1}^{2} + (1 - λ)^{2} σ_{2}^{2}$

Explanation

Find the value of $λ$ yields the estimate having the lowest possible variance? Differentiate with respect to $λ$ and then equating to $0 .$

role="math" $\frac{d}{d λ} [λ^{2} σ_{1}^{2} + (1 - λ)^{2} σ_{2}^{2}] = 0$

$2 λ σ_{1}^{2} - 2 (1 - λ) σ_{2}^{2} = 0$

$λ = \frac{σ_{2}^{2}}{σ_{1}^{2} + σ_{2}^{2}}$

Final Answer

Therefore,

Explain it is desirable to use this value of $λ .$

As $Var [λ X_{1} + (1 - λ) X_{2}] = E [{(λ X_{1} + (1 - λ) X_{2})}^{2}]$ , then $λ$ is to be small.

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