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Let $\{\mu_N\}_N$ be a sequence of $N$ dimensional vectors, and $\{\Lambda_N \}$ a sequence of $N \times N$ symmetric positive definite precision matrices. I drop the $N$ subscripts for simplicity of notation. Assume \begin{equation} \lim_{N \to \infty} \mu' \Lambda \mu = S \end{equation} where $S$ is a positive scalar. Assume $\bar X_T$ (which is the mean of $T$ iid random vectors) has a multivariate normal distribution \begin{equation} \bar X_T \sim N \left( \mu, \frac{1}{T} \Lambda^{-1} \right) \end{equation}

The Question: How do you solve for the following probability limit? \begin{equation} \text{plim}_{\substack{N / T = c \\ N,T \to \infty}} \bar X_T' \Lambda \bar X_T \end{equation} for some positive scalar $c$.

Do the probability limit exist? If so, what is the answer?

Carter D.
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1 Answers1

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I found an answer. Since $\Lambda$ is symmetric and positive definite, the square root $\Lambda^{1 / 2}$ exists and is unique. Define \begin{equation} \bar y_T = \sqrt{T} \Lambda^{1/2} \bar x_T \sim N \left( \sqrt{T} \Lambda^{1/2} \mu, I \right) \end{equation} Then we have \begin{equation} \bar x_T' \Lambda \bar x_T = \frac{1}{T} \bar y_T' \bar y_T \end{equation} Now $\bar y_T' \bar y_T$ actually has a noncentral chi-squared distribution with degrees of freedom $k$ and non-centrality parameter $\lambda$ equal to \begin{equation} k = N, \;\; \lambda = (\sqrt{T} \Lambda^{1/2} \mu)' (\sqrt{T} \Lambda^{1/2} \mu) = T \mu' \Lambda \mu \end{equation} Thus \begin{equation} E[x_T' \Lambda \bar x_T] = E \left[ \frac{1}{T} \bar y_T' \bar y_T \right] = \frac{1}{T} (k + \lambda) = \frac{1}{T} \left( N + T \mu' \Lambda \mu \right) = \frac{N}{T} + \mu' \Lambda \mu \to c + S \end{equation} \begin{equation} \text{Var} [x_T' \Lambda \bar x_T] = \text{Var} \left[ \frac{1}{T} \bar y_T' \bar y_T \right] = \frac{1}{T^2} 2 (k + 2 \lambda) = \frac{2 N}{T^2} + \frac{4 \mu' \Lambda \mu}{T} \to 0 \end{equation} Since the variance collapses to zero, it must be the case that \begin{equation} \text{plim}_{\substack{N / T = c \\ N,T \to \infty}} \bar X_T' \Lambda \bar X_T = S + c \end{equation}

Carter D.
  • 65