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lbvjy [14]
2 years ago
15

How do I solve this?

Mathematics
1 answer:
Bogdan [553]2 years ago
3 0

(i) Each of <em>u</em>, <em>v</em>, and <em>w</em> are vectors in R<em>ⁿ</em>, so they each have size <em>n</em> × 1 (i.e. <em>n</em> rows and 1 column). So <em>u </em>and <em>v</em> both have size <em>n</em> × 1, while <em>w</em>ᵀ has size 1 × <em>n</em>.

<em>M</em> is an <em>n</em> × <em>n</em> matrix, so the matrix <em>A</em> has been partitioned into the blocks

A=\begin{pmatrix}M_{n\times n}&\mathbf u_{n\times 1}\\\mathbf w^\top_{1\times n}&\alpha\end{pmatrix}

where <em>α</em> is a scalar with size 1 × 1. So <em>A</em> has size (<em>n</em> + 1) × (<em>n</em> + 1).

(ii) Multiplying both sides (on the left is the only sensible way) by the given matrix gives

\begin{pmatrix}M^{-1}&\mathbf 0\\-\mathbf w^\top M^{-1}&1\end{pmatrix}\begin{pmatrix}M&\mathbf u\\\mathbf w^\top&\alpha\end{pmatrix}\begin{pmatrix}\mathbf x\\x_{n+1}\end{pmatrix}=\begin{pmatrix}M^{-1}&\mathbf 0\\-\mathbf w^\top M^{-1}&1\end{pmatrix}\begin{pmatrix}\mathbf v\\v_{n+1}\end{pmatrix}

\begin{pmatrix}M^{-1}M&M^{-1}\mathbf u\\-\mathbf w^\top M^{-1}M+\mathbf w^\top&-\mathbf w^\top M^{-1}\mathbf u+\alpha\end{pmatrix}\begin{pmatrix}\mathbf x\\x_{n+1}\end{pmatrix}=\begin{pmatrix}M^{-1}&\mathbf 0\\-\mathbf w^\top M^{-1}&1\end{pmatrix}\begin{pmatrix}\mathbf v\\v_{n+1}\end{pmatrix}

and of course <em>M</em> ⁻¹ <em>M</em> = <em>I</em> (the identity matrix), so

-<em>w</em>ᵀ <em>M</em> ⁻¹ <em>M</em>  + <em>w</em>ᵀ = -<em>w</em>ᵀ + <em>w</em>ᵀ = 0ᵀ (the zero vector transposed)

(iii) Simplifying the system further gives

\begin{pmatrix}I&M^{-1}\mathbf u\\\mathbf 0^\top&-\mathbf w^\top M^{-1}\mathbf u+\alpha\end{pmatrix}\begin{pmatrix}\mathbf x\\x_{n+1}\end{pmatrix}=\begin{pmatrix}M^{-1}&\mathbf 0\\-\mathbf w^\top M^{-1}&1\end{pmatrix}\begin{pmatrix}\mathbf v\\v_{n+1}\end{pmatrix}

\begin{pmatrix}\mathbf x+x_{n+1}M^{-1}\mathbf u\\(\alpha-\mathbf w^\top M^{-1}\mathbf u)x_{n+1}\end{pmatrix}=\begin{pmatrix}M^{-1}\mathbf v\\-\mathbf w^\top M^{-1}\mathbf v+v_{n+1}\end{pmatrix}

So now, setting <em>y</em> = <em>M</em> ⁻¹<em>u</em> and <em>z</em> = <em>M</em> ⁻¹ <em>v</em> gives

\begin{pmatrix}\mathbf x+x_{n+1}\mathbf y\\(\alpha-\mathbf w^\top\mathbf y)x_{n+1}\end{pmatrix}=\begin{pmatrix}\mathbf z\\-\mathbf w^\top \mathbf z+v_{n+1}\end{pmatrix}

Given that <em>α</em> - <em>w</em>ᵀ<em>y</em> ≠ 0, it follows that

x_{n+1}=\dfrac{v_{n+1}-\mathbf w^\top\mathbf z}{\alpha-\mathbf w^\top\mathbf y}

(iv) Combining the result from (iii) with the first row gives

\mathbf x+x_{n+1}\mathbf y=\mathbf z

\mathbf x=\mathbf z-x_{n+1}\mathbf y

\mathbf x=\mathbf z-\dfrac{v_{n+1}-\mathbf w^\top\mathbf z}{\alpha-\mathbf w^\top\mathbf y}\mathbf y

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Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.


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