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Michael Hardy
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Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots m_{1N}\\ m_{21}&1&\cdots m_{2N}\\ \vdots&\vdots&\vdots\\ m_{N1}&\cdots&\cdots 1 \end{matrix} \right] \end{equation}\begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots& m_{1N}\\ m_{21}&1&\cdots& m_{2N}\\ \vdots&\vdots&&\vdots\\ m_{N1}&m_{N2}&\cdots& 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;

Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) d\boldsymbol{u} \end{equation}\begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) \,d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$

Now, if let the $(i,j)th$$(i,j)\text{th}$ element of $\boldsymbol{M}$ be an independent variable a,$a,$ $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.

Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u})d\boldsymbol{u}$$F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u}) \, d\boldsymbol{u}$.

My question is:

Is $F(a)$ a monotonically increasing function of $a$?

Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots m_{1N}\\ m_{21}&1&\cdots m_{2N}\\ \vdots&\vdots&\vdots\\ m_{N1}&\cdots&\cdots 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;

Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$

Now, if let the $(i,j)th$ element of $\boldsymbol{M}$ be an independent variable a, $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.

Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u})d\boldsymbol{u}$.

My question is:

Is $F(a)$ a monotonically increasing function of $a$?

Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots& m_{1N}\\ m_{21}&1&\cdots& m_{2N}\\ \vdots&\vdots&&\vdots\\ m_{N1}&m_{N2}&\cdots& 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;

Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) \,d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$

Now, if let the $(i,j)\text{th}$ element of $\boldsymbol{M}$ be an independent variable $a,$ $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.

Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u}) \, d\boldsymbol{u}$.

My question is:

Is $F(a)$ a monotonically increasing function of $a$?

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Does Is this function is monotonically incresing about aincreasing?

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Does this function is monotonically decreasingincresing about a?

Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots m_{1N}\\ m_{21}&1&\cdots m_{2N}\\ \vdots&\vdots&\vdots\\ m_{N1}&\cdots&\cdots 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;

Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u}) d\boldsymbol{u} \end{equation}\begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\gamma)=\left(-\infty,\gamma\right)\times\cdots\times \left(-\infty,\gamma\right)$$\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$

Now, if let the $(i,j)th$ element of $\boldsymbol{M}$ be an independent variable a, $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.

Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u})d\boldsymbol{u}$.

My question is:

Is $F(a)$ a monotonically decreasingincreasing function of $a$?

Does this function is monotonically decreasing about a?

Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots m_{1N}\\ m_{21}&1&\cdots m_{2N}\\ \vdots&\vdots&\vdots\\ m_{N1}&\cdots&\cdots 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;

Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u}) d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\gamma)=\left(-\infty,\gamma\right)\times\cdots\times \left(-\infty,\gamma\right)$

Now, if let the $(i,j)th$ element of $\boldsymbol{M}$ be an independent variable a, $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.

Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u})d\boldsymbol{u}$.

My question is:

Is $F(a)$ a monotonically decreasing function of $a$?

Does this function is monotonically incresing about a?

Suppose that $\boldsymbol{t}\sim \mathcal{N}(\boldsymbol{u};\boldsymbol{0},\boldsymbol{M})=f_{\boldsymbol{t}}(\boldsymbol{u})$, where $\boldsymbol{t}$ is a $N$-dimensional gaussian random vector, and \begin{equation} \boldsymbol{M}= \left[ \begin{matrix} 1&m_{12}&\cdots m_{1N}\\ m_{21}&1&\cdots m_{2N}\\ \vdots&\vdots&\vdots\\ m_{N1}&\cdots&\cdots 1 \end{matrix} \right] \end{equation} where $m_{i,j}=m_{j,i}\leq1$, $i=1,2,\cdots,N$, and $j=1,2,\cdots,N$;

Let \begin{equation} F=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\boldsymbol{\gamma})} f_{\boldsymbol{t}}(\boldsymbol{u}) d\boldsymbol{u} \end{equation} where $\mathcal{Y}(\boldsymbol{\gamma})=\left(-\infty,\gamma_1\right)\times\cdots\times \left(-\infty,\gamma_N\right)$

Now, if let the $(i,j)th$ element of $\boldsymbol{M}$ be an independent variable a, $i\neq j$, i.e., we have $m_{i,j}=m_{j,i}=a<1$.

Then we let $F(a)=\int\limits_{\boldsymbol{u}\in\mathcal{Y}(\gamma)} f_{\boldsymbol{t}}(\boldsymbol{u})d\boldsymbol{u}$.

My question is:

Is $F(a)$ a monotonically increasing function of $a$?

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