Let $U_1=(X_1,Y_1)^T,\dots,U=(X_n,Y_n)^T$ are i.i.d. copies of $U=(X,Y)^T\sim N_2(0,\Sigma)$ where $$ \Sigma= \begin{pmatrix} \sigma^2 & \rho\sigma\tau \\ \rho\sigma\tau & \tau^2 \end{pmatrix} $$ such that \begin{equation} \sum_{i=1}^n\min\{U_iU_i^T,c\} \end{equation} is invertible where $\min\{A,c\}$ represents the matrix with element $\min\{a_{ij},c\}$. Given $c>0$ and define \begin{equation} b=\frac{\sum_{i=1}^n\max\{-c,\min\{X_iY_i,c\}\}}{\left(\sum_{i=1}^n\min\{X_i^2,c\}\sum_{i=1}^n\min\{Y_i^2,c\}\right)^{1/2}}. \end{equation}

The problem is how to prove that \begin{equation} -E(\min\{\sigma\tau Z^2,c\})\leq b\leq E(\min\{\sigma\tau Z^2,c\}), \end{equation} where $Z$ is normal standard distribution. I don't know how can I start this problem. Any advice?

  • $\begingroup$ Wouldn't you want $c^2$ to appear in the expressions in the denominator rather than $c$? (And perhaps also in the last inequalities?) What is $Z$? $\endgroup$ – whuber Feb 17 '14 at 15:08
  • $\begingroup$ @whuber: Thanks for your comments. I have added what $Z$ is. And for $c$, I would write $c$ indeed since I have stated that $c$ is positive $\endgroup$ – Jlamprong Feb 17 '14 at 15:41

Notice that $$\mathbb{E}(\min\{\sigma\tau Z^2, c\}) \to 0$$ as $\sigma\tau\to 0.$ However, provided both $\sigma$ and $\tau$ are nonzero and $n=2$ it is possible for the outcomes to be $$\{(X_i,Y_i), i=1,2\} = \{\left(2c,\frac{1}{2}\right), \left(2c,\frac{1}{2}\right)\},$$ in which case if $c \ge 1/4,$ then $\max\{-c,\min\{X_iY_i, c\}\}=c,$ $\min\{X_i^2,c\}=c,$ and $\min\{Y_i^2,c\}=1/4,$ entailing $$b = \frac{c + c}{\sqrt{(c+c)(\frac{1}{4}+\frac{1}{4})}} = 2\sqrt{c}.$$ Since there is no assumed relationship among $c,\sigma,\tau,$ and $\rho,$ choose $\sigma$ and $\tau$ sufficiently small to ensure that $\mathbb{E}(\min\{\sigma\tau Z^2, c\}) \lt 2\sqrt{c},$ contradicting what you are trying to prove.

The basic problem is that you are attempting to prove an inequality about arbitrary real numbers as expressed by the formula for $b$; their distribution--provided it is supported on the entire real line--is irrelevant.

Response to edits

The condition was added in a later edit. Note, though, that in the counterexample

$$\min\{U_iU_i^\prime,c\} = \min\{\pmatrix{4c^2 & c \\ c & \frac{1}{4}}, c\} = \pmatrix{c & c \\ c & \frac{1}{4}}$$

(because $c\gt \frac{1}{4}$). Their sum is equal to $\pmatrix{2c & 2c \\ 2c & \frac{1}{2}}$ whose determinant $c - 4c^2$ is nonzero, showing the sum is invertible, whence the counterexample still applies.

  • $\begingroup$ Thanks @whuber for your counterexample. It turned out that I forgot the other assumptions. Is that still incorrect? $\endgroup$ – Jlamprong Feb 17 '14 at 16:22
  • $\begingroup$ The edit you made to your question is meaningless: I believe you may have forgotten to include some part of it. Note, too, that the expression $U_iU_i^\prime$ is a two-by-two matrix, calling into question what "min" means. $\endgroup$ – whuber Feb 17 '14 at 16:23
  • $\begingroup$ OK @Whuber. Thanks for your correction. Now, I have added the missing assumption. $\endgroup$ – Jlamprong Feb 17 '14 at 16:30
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    $\begingroup$ Your latest edit does not rescue anything, because in the counterexample for $c \gt 1/4$ the matrix is clearly invertible (its determinant is $c-4c^2\ne 0$). I would like to suggest you let this rest, instead of applying a series of quick edits; rethink your question; and pose a new one when you have better determined what it is you need to show. $\endgroup$ – whuber Feb 17 '14 at 16:35
  • $\begingroup$ OK, Thank you very much for your suggestions. I'll think this problem carefully. $\endgroup$ – Jlamprong Feb 17 '14 at 16:37

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