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 March 6th, 2019, 12:28 PM #1 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 Quadratic forms Suppose that $X \sim N(\mu, \Lambda)$ with det$(\Lambda) > 0$. Show that $Q=(X-\mu)'\Lambda^{-1}(X-\mu)$ has a $\chi^2(n)$ distribution where $n$ is the dimension of $X$. Here is my proof: Since det$(\Lambda) > 0$ then $\Lambda$ is positive definite, $\Lambda=T\Sigma T'$ where $T$ is a real orthogonal matrix and $\Sigma = \begin{bmatrix} \lambda_{1} &0 &\cdots &0 \\ 0 &\lambda_{2} &\cdots &0 \\ \vdots &\vdots &\ddots &\vdots \\ 0 &0 &\cdots &\lambda_{n} \end{bmatrix} \quad {.Then }$ $\Lambda^{-1} = T\Sigma^{-1} T'$. Then we have: \begin{aligned} Q &= (X-\mu)'\Lambda^{-1}(X-\mu) \\ &= (X-\mu)'T\Sigma^{-1} T'(X-\mu) \\ &= Y'\Sigma Y \end{aligned} where $Y=T'(X-\mu)$. Further, \begin{aligned} Q &= Y'\Sigma Y \\ &= \begin{bmatrix} Y_1 & Y_2 &\cdots &Y_n \end{bmatrix} \begin{bmatrix} \dfrac{1}{\lambda_{1}} &0 &\cdots &0 \\ 0 &\dfrac{1}{\lambda_{2}} &\cdots &0 \\ \vdots &\vdots &\ddots &\vdots \\ 0 &0 &\cdots &\dfrac{1}{\lambda_{n}} \end{bmatrix} \begin{bmatrix} Y_1 \\ Y_2 \\ \vdots \\ Y_n \end{bmatrix} \\ &= \sum^{n}_{i=1} \dfrac{Y_i^2}{\lambda_i} = \sum^{n}_{i=1}\left(\dfrac{Y_i}{\sqrt{\lambda_{i}} }\right)^2 \end{aligned} Since $Y_i \sim N(0, \lambda_{i})$ and $X_i$ is independent then $Q= \sum^{n}_{i=1}\left(\dfrac{Y_i^2}{\sqrt{\lambda_{i }}}\right)^2 \sim \chi^2(n).$ There are two other problems that I do not know where to start, can someone give me a hint. Problem 1) Let $X_1, X_2, \cdots, X_n$ be i.i.d. random variables with corresponding order statistics $X_{(1)}, \cdots, X_{(n)}.$ a) Show that $P(X_1=X_{(k)}) = 1/n$ for all $k=1,\cdots,n$. b) Compute $E(X_1|X_{(1)},X_{(2)}, \cdots, X_{(n)})$. Problem 2) Suppose that $X_1, X_2, \cdots, X_n$ are i.i.d. Bernoulli(p) random variables. What is the generating moment function of $X_1+X_2^2+X_3^3+\cdots+X_n^{n}$. Then identify the distribution. Thank you. Last edited by Shanonhaliwell; March 6th, 2019 at 01:03 PM. March 6th, 2019, 03:27 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,533 Thanks: 1390 1a) I believe the idea is that $P[X_1 = X_{(k)}] = \dfrac{(n-1)!}{n!} = \dfrac 1 n$ Think about the ways that the $X_k$ might be sorted b) you're given all the order statistics. (a) showed they form a discrete uniform distribution. So what's the expectation? Thanks from Shanonhaliwell March 6th, 2019, 06:41 PM #3 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 Thank you. I think the expectation is: $E(X_1|X_{(1)},X_{(2)},\cdots,X_{(n)})=\dfrac{1}{n} \sum_{k=1}^{n}X_k$ How about problem 2, any idea. March 6th, 2019, 07:55 PM   #4
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Quote:
 Originally Posted by Shanonhaliwell Since det$(\Lambda) > 0$ then $\Lambda$ is positive definite
The rest of your proof seems fine assuming this is true. However, I don't see why this follows. Is there some additional structure on $\Lambda$ which you haven't mentioned? In general, if a matrix has positive determinant it doesn't mean all of its eigenvalues are positive. It just means it must have an even number of negative eigenvalues.

If this is indeed a mistake, I think this approach can be fixed by using the singular value decomposition for $\Lambda$ in place of its eigenvalue decomposition in your current proof. March 6th, 2019, 09:41 PM   #5
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 Originally Posted by SDK The rest of your proof seems fine assuming this is true. However, I don't see why this follows. Is there some additional structure on $\Lambda$ which you haven't mentioned? In general, if a matrix has positive determinant it doesn't mean all of its eigenvalues are positive. It just means it must have an even number of negative eigenvalues. If this is indeed a mistake, I think this approach can be fixed by using the singular value decomposition for $\Lambda$ in place of its eigenvalue decomposition in your current proof.
$\Lambda$ is a covariance matrix and thus positive semi-definite.

Since $|\Lambda| > 0$ it's positive definite. March 7th, 2019, 03:00 AM   #6
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 Originally Posted by romsek $\Lambda$ is a covariance matrix and thus positive semi-definite. Since $|\Lambda| > 0$ it's positive definite.
This clears it up thanks. March 7th, 2019, 08:52 AM #7 Member   Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 There is a mistake in my proof so I hope someone can fix for it me or let me have a chance to edit my post. It should be: \begin{aligned} Q &= (X-\mu)'\Lambda^{-1}(X-\mu) \\ &= (X-\mu)'T\Sigma^{-1} T'(X-\mu) \\ &= Y'\Sigma^{-1} Y \end{aligned} so \begin{aligned} Q &= Y'\Sigma^{-1} Y \\ &= \begin{bmatrix} Y_1 & Y_2 &\cdots &Y_n \end{bmatrix} \begin{bmatrix} \dfrac{1}{\lambda_{1}} &0 &\cdots &0 \\ 0 &\dfrac{1}{\lambda_{2}} &\cdots &0 \\ \vdots &\vdots &\ddots &\vdots \\ 0 &0 &\cdots &\dfrac{1}{\lambda_{n}} \end{bmatrix} \begin{bmatrix} Y_1 \\ Y_2 \\ \vdots \\ Y_n \end{bmatrix} \\ &= \sum^{n}_{i=1} \dfrac{Y_i^2}{\lambda_i} = \sum^{n}_{i=1}\left(\dfrac{Y_i}{\sqrt{\lambda_{i}} }\right)^2 \end{aligned} For the second problem, since $X_1, X_2, \cdots, X_n$ are i.i.d. Bernoulli(p) random variables so $X_2^2=X_2,X_3^3=X_3,\cdots,X_n^{n}=X_n$ then $\sum^{n}_{i=1}X_{i}^{i}=\sum_{i=1}^{n}X_i \sim \text{Binomial distribution} (n,p)$. Hence the generating moment function of $X_1+X_2^2+X_3^3+\cdots+X_n^{n}$ is the generating moment function of Binomial distribution which is $(1-p+pe^t)^n$. Thank you @romsek and @SDK for your helps. Last edited by Shanonhaliwell; March 7th, 2019 at 09:24 AM. Tags forms, quadratic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Luiz Linear Algebra 4 November 3rd, 2015 07:17 AM Akcope Calculus 0 March 20th, 2014 08:43 AM !_UK@\$ Elementary Math 2 August 13th, 2010 12:04 AM antonio85 Number Theory 0 July 15th, 2010 07:28 AM depressedmaths Differential Equations 0 June 8th, 2009 12:30 AM

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