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.

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