10 6 3 Positively Correlated Normal Test Statistics

1). We want to find an expression of the conditional distribution of $\hat{S}_{-i}$ given $\hat{S}_{i}$. Writing $\Sigma=\begin{pmatrix} \Sigma_{-i,-i} & \Sigma_{i,-i} \\ \Sigma_{-i,i} & \Sigma_{i,i} \end{pmatrix}$ and $(\Sigma)^{-1}=K=\begin{pmatrix} K_{-i,-i} & K_{i,-i} \\ K_{-i,i} & K_{i,i} \end{pmatrix}$.

According to the Lemma A.4 in appendix A this conditional distribution is $\mathcal{N}(-{(K_{-i,-i})}^{-1}K_{i,-i}\hat{S}_{i},K_{-i,-i})$.

Starting from the equation : $\Sigma \cdot K = Id_{m}$ we find four equations

(1)

\begin{eqnarray} \Sigma_{-i,-i}K_{-i,-i}+\Sigma_{i,-i}K_{-i,i}&=&Id_{m-1}\\ \Sigma_{i,i}K_{i,-i}+\Sigma_{i,-i}K_{i,i}&=&\begin{pmatrix}0\\\vdots\\0\end{pmatrix} \\ \Sigma_{-i,i}K_{i,i}+\Sigma_{i,i}K_{-i,i}&=&\begin{pmatrix}0 & \cdots &0\end{pmatrix} \\ \Sigma_{-i,i}K_{i,-i}+\Sigma_{i,i}K_{i,i}&=& 1 \end{eqnarray}

Then, using the third and the first equations we find the covariance matrix of the conditional distribution :

(2)

\begin{align} (K_{-i,-i})^{-1}=\Sigma_{-i,-i}-\Sigma_{i,-i}\Sigma_{-i,i}(\Sigma_{i,i})^{-1} \end{align}

Finally, using the formula of $(K_{-i,-i})^{-1}$,

(3)

\begin{align} -(K_{-i,-i})^{-1}K_{i,-i}=-[\Sigma_{-i,-i}-\Sigma_{i,-i}\Sigma_{-i,i}(\Sigma_{i,i})^{-1}]K_{i,-i} \end{align}

after developping and using the second and fourth equation we find,

(4)

\begin{eqnarray} -(K_{-i,-i})^{-1}K_{i,-i}&=&-\Sigma_{i,-i}K_{i,i}+\frac{1}{\Sigma_{i,i}}\Sigma_{i,-i}(1-\Sigma_{i,i}K_{i,i})\\ &=& \frac{1}{\Sigma_{i,i}}\Sigma_{i,-i} \end{eqnarray}

We conclude by noticing that in the Lemma A.4 hypothesis, random variables have a mean equal to 0. Then,

(5)

\begin{align} -(K_{-i,-i})^{-1}K_{i,-i} =\mu_{-i} + \frac{1}{\Sigma_{i,i}}\Sigma_{i,-i}(\hat{S}_{i}-\mu_{i}) \end{align}

2.) We remember that $T_{i}(x)=\mathbb{P}(\epsilon_{i}\geqslant x)$.
Let's define the function ${T_{i}}^{-1}(u)=\inf{ \{s \in \mathbb{R}: T_{i}(s)\leqslant u\} }$. We want to prove that: $\{ T_{i}(s)\leqslant u \}=\{ {T_{i}}^{-1}(u)\leqslant s \}$, $\forall$ $(u,s) \in [0,1]\times\mathbb{R}$.

We define $I=\{ s \in \mathbb{R}, T_{i}(s)\leqslant u \}$, $u \in [0,1]$.
Because $\lim_{s:+\infty}T_{i}(s)=0$ and $\lim_{s:-\infty}T_{i}(s)=1$ we have $I\neq \emptyset$. Furthermore, for $t_{1} \in I$ and $t_{2}\geqslant t_{1}$, $t_{2}$ is in $I$ because $T_{i}$ is nonincreasing.
Then there is a $\alpha \in \mathbb{R}$ such that $I=[\alpha,+\infty[$ or $I=]\alpha,+\infty[$.
As $T_{i}$ is "cad-lag", so ${T_{i}}^{-1}$ is also "cad-lag".
This allows us to say that $I=[\alpha,+\infty[$. And finally that for $u \in [0,1]$

(6)

\begin{align} I=\{ s \in \mathbb{R}, T_{i}(s)\leqslant u \}=[{T_{i}}^{-1}(u),+\infty[. \end{align}

Hence the results.

For $i \in \{ 1,...m \}$ and $u \in [0,1]$. Noticing that

(7)

\begin{eqnarray} \mathbb{E}[g(\hat{p}_{1},...,\hat{p}_{m})|\hat{p}_{i}\leqslant u] &=& \mathbb{E}[g(T_{1}(\hat{S}_{1}),...,T_{m}(\hat{S}_{m})) | T_{i}(\hat{S}_{i})\leqslant u], \\ &=& \mathbb{E}[f(\hat{S}_{i},\hat{S}_{-i})|T_{i}(\hat{S}_{i})\leqslant u]. \end{eqnarray}

We are finally allowed to say that

(8)

\begin{align} \mathbb{E}[g(\hat{p}_{1},...,\hat{p}_{m})|\hat{p}_{i}\leqslant u]=\mathbb{E}[f(\hat{S}_{i},\hat{S}_{-i})| \hat{S}_{i}\geqslant {T_{i}}^{-1}(u)]. \end{align}

With f, mesurable and bounded because of g, nonincreasing by composition of g nondecreasing and of $T_{1},...,T_{m}$ each nonincreasing.

3.) Using the previous question

(9)

\begin{eqnarray} \mathbb{E}[g(\hat{p}_{1},...,\hat{p}_{m})|\hat{p}_{i}\leqslant u] &=& \mathbb{E}[f(\hat{S}_{i},\hat{S}_{-i})| \hat{S}_{i}\geqslant {T_{i}}^{-1}(u)], \\ &=& \mathbb{E}[\mathbb{E}[\underbrace{f(\hat{S}_{i},\hat{S}_{-i}) | \hat{S}_{i}]}_\textrm{$=\Phi (\hat{S}_{i})$ } | \hat{S}_{i}\geqslant {T_{i}}^{-1}(u)], \\ &=& \mathbb{E}[\Phi(\hat{S}_{i}) | \hat{S}_{i}\geqslant {T_{i}}^{-1}(u)], \\ &\overset{\Phi \geq 0}{=}& \int_{0}^{+\infty} \mathbb{P}(\Phi (\hat{S}_{i})\geqslant t | \hat{S}_{i}\geqslant {T_{i}}^{-1}(u) ) dt. \\ \end{eqnarray}

4.)
We want to prove that the function $h: u \mapsto \mathbb{E}[g(\hat{p}_{1},...,\hat{p}_{m})|\hat{p}_{i}\leqslant u]$ is nondecreasing for any $i \in \{1,...m\}$ and any function $g:[0,1]^m \mapsto \mathbb{R}^{+}$ mesurable, bounded and nondecreasing.

The function $\Phi$ is nonincreasing, mesurable, bounded and positive because f is nonincreasing, mesurable, bounded and positive.

Then, using the previous question,

(10)

\begin{eqnarray} \int_{0}^{+\infty} \mathbb{P}(\Phi (\hat{S}_{i})\geqslant t | \hat{S}_{i}\geqslant {T_{i}}^{-1}(u) ) dt &=& \int_{0}^{+\infty} \mathbb{P}( \Phi(\hat{S}_{i}) \geqslant t | \Phi(\hat{S}_{i})\leqslant \Phi ({T_{i}}^{-1}(u)) ) dt. \\ \end{eqnarray}

To conclude, we simply notice that

(11)

\begin{align} u \mapsto \mathbb{P}( \Phi(\hat{S}_{i}) \geqslant t | \Phi(\hat{S}_{i})\leqslant \Phi ({T_{i}}^{-1}(u)) ) = (1-\frac{\mathbb{P}(\Phi(\hat{S}_{i}) \leqslant t)}{\mathbb{P}(\Phi(\hat{S}_{i})\leqslant \Phi ({T_{i}}^{-1}(u)))})_{+} \end{align}

is nondecreasing for all $t \in \mathbb{R}^{+}$ (using that $u \mapsto {T_{i}}^{-1}(u)$ is nonincreasing).

Solution proposed by Tabouy T.