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Since I do not have an answer on Quantitative Finance in my question I cross-post here the problem to tag some other categories

The following assumptions are part of the paper of Back, Chao and Willard and I can not solve for the statistic that is denoted as $\phi$ in the sequel. I would be glad if anyone could help me. Below i set the assumptions and the equations of interest

Suppose that in the market, there are $N\geq 1$ informed agents, who trade a risky asset continuously in the time interval $[0,1)$. Each agent $i$ receives a mean-zero signal $\tilde{s}^i$ at time 0. We assume the signals and the liquidation value of the asset have a nondegenerate joint normal distribution that is symmetric in the signals. Symmetry means that the joint distribution of the asset value and the signals $\tilde{s}^1,...,\tilde{s}^N$ is invariant to a permutation of the indices $1,...,N$. Let $\tilde{v}$ denote the expectation of the liquidation value conditional on the combined information of the informed traders. By normality, $\tilde{v}$ is an affine function of the $\tilde{s}^i$. By rescaling the $\tilde{s}^i$ if necessary, we can assume without loss of generality that

\begin{equation}\tilde{v}=\bar{v}+\Sigma^{N}_{i=1} s^i\end{equation} for a constant $\bar{v}$. For simplicity, we assume $\bar{v}=0$. Let \begin{align}\phi=\frac{var(\tilde{v})}{var(N\tilde{s}^i)}\end{align}

The statistic $\phi$ is a measure of the quality of each agent’s information. Specifically, it is the $R^2$ in the linear regression of $\tilde{v}$ on $\tilde{s}^i$, that is, it is the percentage of the variance in $\tilde{v}$ that is explained by the trader’s information.

Letting $\rho$ denote the correlation coefficient of $\tilde{s}^i$ with $\tilde{s}^j$ for $i\neq j$, one can compute $\phi$ for $N>1$ as

\begin{equation}\phi=\frac{1}{N}+\frac{N-1}{N}\rho\end{equation}

If $\phi=1$, then either $N=1$ or the $\tilde{s}^i$ are perfectly corellated. In either case each informed trader has perfect information about $\tilde{v}$.

My questions are the following

  1. what does it mean intuitively "a nondegenerate joint normal distribution" and in particular I would like to understand the term nondegenerate.
  2. What does it mean "invariant to indices" ?
  3. the liquidation value is equal to the sume of the signals, does this come from the assumption that it it an affine function of the $\tilde{s}^i$?
  4. How do we find that measure $\phi$? is it from the linear regression of $\tilde{v}$ on $\tilde{s}^i$?
  5. How $\phi$ is tranformed to \begin{equation}\phi=\frac{1}{N}+\frac{N-1}{N}\rho\end{equation}

Here it is a link from the paper

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  • $\begingroup$ @1muflon1♦ Thank you! $\endgroup$
    – Nav89
    Nov 25, 2020 at 18:37
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    $\begingroup$ you are welcome - it is important to make distinction between reposting and cross posting on different sites some might have not checked the link and just downvoted/voted to close $\endgroup$
    – 1muflon1
    Nov 25, 2020 at 18:39

2 Answers 2

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  1. A degenerate joint normal is distribution is one in which you cannot find a PDF for the distribution. They assume you can. (The covariance matrix is invertible).

  2. Let $f(s_1,s_2\dots,s_n,v)$ be the distribution. If I was to exchange $s_1$ for $s_2$, $f(s_2,s_1,\dots,s_n,v) = f(s_1,s_2\dots,s_n,v)$, the distribution does not change. And you can exchange as many indices (signals) as you'd like.

  3. The assumption that the distribution is jointly normal implies $\tilde{v}$ is an affine function of $\tilde{s}_i$, which then implies the sum of signals.

  4. Regress $\tilde{v}$ on $\tilde{s}_i$ and calculate $R^2$ (the Coefficient of determination).

  5. This is something I hope someone else can answer.

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  • $\begingroup$ Ok I know about 4. and 5. but could you explain me, where this $N$ in the denominator of the $\phi$ comes from? \begin{align}\phi=\frac{var(\tilde{v})}{var(N\tilde{s}^i)}\end{align} Could you provide some calculations. I am confused. $\endgroup$
    – Nav89
    Nov 25, 2020 at 20:02
  • $\begingroup$ Also,could the symmetric joint normal distribution assumption for the informed trader's signals mean the covariance bwteen the invividula singals and the asset valuation is identical? $\endgroup$
    – Nav89
    Nov 26, 2020 at 12:31
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Well, I will try to answer 4.

We know that the asset liquidation value $\tilde{v}$ is an affine function of the singals thus we have that $$\tilde{v}=\bar{v}+\sum_{i=1}^{N}\tilde{s}^i\Rightarrow \tilde{v}=\bar{v}+N\underbrace{\frac{\sum_{i=1}^{N}\tilde{s}^i}{N}}_{\tilde{s}^i}\Rightarrow\tilde{v}=\bar{v}+N\tilde{s}^i$$ where the $\tilde{s}^i$ is the average singal that is a sufficient statistic to infer the liquidation value of the asset conditioning on it instaed of the individual signal since this is also driven by the assumption that the signals and the liquidation value of the asset have a nondegenerate joint normal distribution that is symmetric in the signals. Hence the expectation of the liquidation value conditional on the combined information of the informed traders is given by the projection theorem to be (projecting $\tilde{s}^i$ on $\tilde{v}$):

$$\mathbb{E}[\tilde{v}|\tilde{s}^i]=\mathbb{E}[\tilde{v}]+\frac{\mathbb{C}ov(\tilde{v},\tilde{s}^i)}{\mathbb{V}ar(\tilde{s}^i)}\left(\tilde{s}^i-\mathbb{E}(\tilde{s}^i)\right)\Rightarrow\mathbb{E}[\tilde{v}|\tilde{s}^i]=\bar{u}+\frac{\mathbb{C}ov(\tilde{v},(\tilde{v}-\bar{v})/N)}{\mathbb{V}ar(\tilde{s}^i)}\tilde{s}^i\Rightarrow\\ \mathbb{E}[\tilde{v}|\tilde{s}^i]=\bar{v}+\frac{\mathbb{V}ar(\tilde{v})}{N^2\mathbb{V}ar(\tilde{s}^i)}\sum_{i=1}^{N}\tilde{s}^i\Rightarrow\mathbb{E}[\tilde{v}|\tilde{s}^i]=\bar{v}+\underbrace{\frac{\mathbb{V}ar(\tilde{v})}{\mathbb{V}ar(N\tilde{s}^i)}}_{\beta^{i}}\sum_{i=1}^{N}\tilde{s}^i$$

where $\beta^{i}$ denotes the beta coefficient of the linear regression of $\tilde{v}$ on $\tilde{s}^i$, that coincide with the $R$-square coefficient and as a consequence

$$\phi=\frac{\mathbb{V}ar(\tilde{v})}{\mathbb{V}ar(N\tilde{s}^i)}$$

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    $\begingroup$ Well, it seems to be ok...but let us see what others can tell about this...and then i will endorse the answer, since I am not a specialist. Anyways, thanks... $\endgroup$
    – Nav89
    Nov 26, 2020 at 14:26
  • $\begingroup$ You're welcome...I am not a specialist either... $\endgroup$ Nov 26, 2020 at 14:27
  • $\begingroup$ just one more thing...could you post your answer here quant.stackexchange.com/questions/59483/… as well. $\endgroup$
    – Nav89
    Nov 26, 2020 at 14:39
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    $\begingroup$ I make the same comment here... Well, if you read this paper, that is mentioned in the paper you are posting onlinelibrary.wiley.com/doi/epdf/10.1111/… maybe you can find out how this $\phi$ is transformed to 5. I think that it has to do, with the conditional expectationg of the signals of the $j^{th}$ and the $i^{th}$ trader...namely try to find this $$\mathbb{E}[\tilde{s}^j|\tilde{s}^i]=\mathbb{E}[\tilde{s}^j]+\frac{\mathbb{C}ovar(\tilde{s}^j,\tilde{s}^i)}{\mathbb{V}ar(\tilde{s}^i)}\left(\tilde{s}^i-\mathbb{E}(\tilde{s}^i)\right)$$ $\endgroup$ Nov 28, 2020 at 15:18

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