Scenario 1: Let $X_i,i\in I$ where $I$ is the index set. For simplicity let $I$ be the finite set $I=\{1,2,\ldots,n\}$. Let each $X_i$ be independent and distributed as $$ X_i\sim N(\delta,1), $$ and let $\mathbf{X}=(X_1,X_2,\ldots,X_n)$. Define the index estimator $$ \gamma(\mathbf{X})=\text{argmax}_{i\in I}{\mathbf{(X)}} $$ The distribution of $\gamma(\mathbf{X})$ is uniform. This is easy to see because each $X_i$ is equally likely to be the maximum since they are all independent and identically distributed.

Scenario 2: In this scenario, $X_i$ are still independent, however, not identically distributed. Each $X_i$ has the distribution $$ X_i\sim N(\delta_i,1), $$ and $\gamma(\mathbf X)$ has the same definition, i.e., $$ \gamma(\mathbf{X})=\text{argmax}_{i\in I}{\mathbf{(X)}}. $$ My question is what is the distribution of $\gamma(\mathbf{X})$ in this scenario? I started with \begin{eqnarray} \Pr(\gamma(\mathbf X)=i)&=&\Pr(X_i=\max\{X_1,X_2,\ldots,X_n\})\\ &=& \Pr(X_i\geq\max\{X_k:k\neq i,k=1,2,\ldots,n\})\\ \end{eqnarray} And define $Y=\max\{X_k:k\neq i,k=1,2,\ldots,n\}$ which implies $$ \Pr(Y\leq y)=\prod_{k \neq i}\Pr(X_k\leq y)=\prod_{k \neq i}\Phi(y) $$ Then my problem simplifies to $$ \Pr(\gamma(\mathbf X)=i)=\Pr(X_i-Y\geq 0). $$ It is the distribution of $X_i-Y$ that makes life complicated for me. However, if you can proceed in another way it might be easier.


For $n = 2$, $p(\gamma(x) = 1) = p(x_{1} - x_{2} > 0)$. Let $z_{1,2} = x_{1} - x_{2}$. Obviously, $z_{1,2} \sim N(\delta_{1} - \delta_{2}, 2)$. Hence, $p(\gamma(x) = 1) = 1 - \Phi((\delta_{1} - \delta_{2}) / \sqrt{2})$.

For $n > 2$, we use multivariate normal distribution. Take $n = 3$ for example. $p(\gamma(x) = 3) = p(x_{3} - x_{2} > 0, x_{3} - x_{1} > 0)$. Let $z_{3,2} = x_{3} - x_{2}$ and $z_{3,1} = x_{3} - x_{1}$. Of course, the vector $(z_{3,2}, z_{3,1})$ follows bivariate normal: \begin{equation*} (z_{3,2}, z_{3,1})' \sim N \left( (\delta_{3} - \delta_{2}, \delta_{3} - \delta_{1})', \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \right) \end{equation*} Then the probability of $p(\gamma(x) = 3) = p(z_{3,2} > 0, z_{3,1} > 0)$ can be calculated from the CDF of the above bivariate normal distribution. The general case of $n > 2$ is similar. You will define a $n-1$-dimensional multivariate normal, whose covariance matrix has diagonal elements of $2$ and off-diagonal elements of $1$.

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  • $\begingroup$ Something counter-intuitive is going, however, consider $n=2$ and if $\delta_1>\delta_{2}$, I will expect $p(\gamma(x)=1)>p(\gamma(x)=2)$, somehow the reverse is true. What are we missing? $\endgroup$ – Chamberlain Foncha Mar 8 '18 at 17:31
  • $\begingroup$ Ok I see the error you omitted a negative sign. $\endgroup$ – Chamberlain Foncha Mar 8 '18 at 18:13

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