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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 $E(X_1X_{(1)},X_{(2)}, \cdots, X_{(n)})$. 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 $P[X_1 = X_{(k)}] = \dfrac{(n1)!}{n!} = \dfrac 1 n$ 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? 
March 6th, 2019, 06:41 PM  #3 
Member Joined: Feb 2018 From: Canada Posts: 46 Thanks: 2 
Thank you. I think the expectation is: \[ E(X_1X_{(1)},X_{(2)},\cdots,X_{(n)})=\dfrac{1}{n} \sum_{k=1}^{n}X_k \] How about problem 2, any idea. 
March 6th, 2019, 07:55 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,533 Thanks: 1390  Quote:
Since $\Lambda > 0$ it's positive definite.  
March 7th, 2019, 03:00 AM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics  
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 $(1p+pe^t)^n$. Thank you @romsek and @SDK for your helps. Last edited by Shanonhaliwell; March 7th, 2019 at 09:24 AM. 

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