Hilbert-Schmidt determinant

Posted on November 23, 2014 by What's up?

In the previous post, I mentioned the intuition of Fredholm determinant for trace class operator, which is $Tr(K)<\infty$ . When it comes to Hilbert-Schmidt determinant, which is $Tr(K^*K)<\infty$ , that definition(let’s call it $det_1$ ) may fail since trace of H-S operator my be infinity. So we need to come up with another definition(let’s call it $det_2$ ) for H-S operator that makes sense.

The intuition is that consider the eigen values of a H-S matrix $K$ , say $\lambda_1,\lambda_2,dots$ ,notice that for $\forall \lambda, 1+\lambda\leq exp(\lambda)$ , so we can define $det_2(I+K)=\prod(\frac{1+\lambda_i}{exp(\lambda_i)})=det_1(I+K)\cdot exp(-Tr(K))$

Fredholm determinant

Posted on November 21, 2014 by What's up?

In the study of measure transformation in Gaussian space, there is a fundamental issue that I want to write down here, which is Fredholm determinant. Let me try to use half an hour to explain the intuition of this determinant in matrix form with finite dimension.

So suppose a matrix $A=(a_{ij})\in \mathbb{R}_{n\times n}$ , our goal is to find out what definition $det(I+A)$ is that makes sense for trace class operator(i.e. $Tr(A)<\infty$ ). We know that its trace $Tr(A)=\sum a_{ii}$ , and assume the eigenvalues of $A$ are $\lambda_1,\lambda_2, \dots, \lambda_n$ , we have

$AX=\begin{pmatrix} \lambda_1 & & &\\ & \lambda_2 & 0 \\ & & \ddots &\\ & & & \lambda_n \end{pmatrix} X$ , so

$Tr(A)=Tr(X^{-1}AX)=\sum \lambda_i$ .

And because $det(I+A)=\prod (1+\lambda_i)=\sum_{k=0}^{n}(\sum_{i_1,\dots,i_k}\lambda_{i_1}\dots\lambda_{i_k})$

Iet’s take a look at these terms in the summation,

when $k=0$ , it’s 1,

when $k=1$ , it’s $\lambda_1+\lambda_2+\dots+\lambda_n$ ,

when $k=2$ , it’s $\sum_{i\neq j}\lambda_i\lambda_j$ ,

which is the trace of operator $\Lambda^2(A)$ on a linear space with basis $e_i\wedge e_j, i<j$ where is $\wedge$ is the wedge product form.

Similarly, we have the expression given in widipedia about “Fredholm determinant” that for a general trace-class operator $A$

$det(I+A)=\sum_{k=0}^{\infty}Tr(\Lambda^k(A))$ and this new operator $\Lambda^k(A)$ is a linear operator on space formed by the basis $\{e_{i_1}\wedge e_{i_2}\wedge\dots\wedge e_{i_k}| i_1<i_2<\dots<i_k\}$

Finally, I want to mention that when $Tr(A)=\sum |\lambda_i|<\infty$ , easy to check that $det(A)<\infty$ indeed.

One way ANOVA vs Two way ANOVA

Posted on November 11, 2014 by What's up?

Today in class we talked about Two-way ANOVA, so my question is how to understand the essence of these two kinds of ANOVA? I’ll try to spend less than one hour to write down what I thought about it.

Let’s take a look at an example, we have treatments $i \in \{1,2,\dots,I\}$ and groups $j\in \{1,2,\dots,J\}$ and for each treatment $i$ and group $j$ , we have $m_{i,j}$ samples, so our two-way ANOVA model should would be

$\mu_{ijk}=\mu_{ij}+\epsilon_{ij;k}$ where $\mu_{ij}$ is the mean of the $i$ th treatment and $j$ th group.

What if we want to use one-way ANOVA to model it? What would that be look like? Because there may be interaction between the treatment and group, so in my understanding, we should consider this model be a linear model with respect to three terms, i.e. treatment level , group level and interaction level , so the model should be $X_{ijk}=\mu_0+\alpha_{i} +\beta_{j} +\gamma_{ij} +\epsilon_{ijk}$ , now we consider the least square solution for $\mu_0,\alpha_i,\beta_j,\gamma_{ij}$ we have

$\min_{\mu_0,\alpha_i,\beta_j,\gamma_{ij}} S=\sum_{i,j,k} (X_{ijk}-\mu_0-\alpha_i-\beta_j-\gamma_{ij})^2$

By taking derivatives with respect to $\mu_0,\alpha,\beta,\gamma$ , we have

$\mu_0 \sum_{ij} m_{ij}=\sum_{ijk} X_{ijk}-\sum_{i}\alpha_i(\sum_j m_{ij})-\sum_{j}\beta_j(\sum_{i} m_{ij})-\sum_{i,j} \gamma_{ij} m_{ij}$

$\alpha_i\sum_j m_{ij}=\sum_{jk} X_{ijk}-\mu_0\sum_j m_{ij}-\beta_i\sum_j m_{ij}-\sum_j(\gamma_{ij} m_{ij})$

$\beta_j\sum_j m_{ij}=\sum_{ik} X_{ijk}-\mu_0\sum_i m_{ij}-\alpha_j\sum_i m_{ij}-\sum_i(\gamma_{ij} m_{ij})$

$\gamma_{ij}=\frac{1}{m_{ij}}\sum_k X_{ijk}-\mu_0 -\alpha_i-\beta_j$

Then the solution of these equations is the two-way ANONA estimator.

So as we can see that if there is only one factor, there would not exist the interaction term $\gamma$ , then we can just regard the original linear model be $I$ different group and analyse individually.

So in sum, so called “Two-way ANOVA” is mainly dealing with the interaction between two different “factors”. So question is if there are three different factors, can we follow the same idea do the “many-factors ANOVA”?

VictorX

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Monthly Archives: November 2014

Hilbert-Schmidt determinant

Fredholm determinant

One way ANOVA vs Two way ANOVA