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 Shanonhaliwell March 6th, 2019 12:28 PM

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.

 romsek March 6th, 2019 03:27 PM

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?

 Shanonhaliwell March 6th, 2019 06:41 PM

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.

 SDK March 6th, 2019 07:55 PM

Quote:
 Originally Posted by Shanonhaliwell (Post 606554) 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.

 romsek March 6th, 2019 09:41 PM

Quote:
 Originally Posted by SDK (Post 606566) 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.

 SDK March 7th, 2019 03:00 AM

Quote:
 Originally Posted by romsek (Post 606567) $\Lambda$ is a covariance matrix and thus positive semi-definite. Since $|\Lambda| > 0$ it's positive definite.
This clears it up thanks.

 Shanonhaliwell March 7th, 2019 08:52 AM

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.

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