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I am currently trying to learn the two related concepts of the Rao-Blackwell theorem and the Lehmann-Scheffé theorem. My question relates to example 2.16 from this document, and I also found much relevant information in this document.

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. My understanding is that, using Lehmann-Scheffé, we get that $E[S^2 \mid \bar{X}] = \bar{X}$. Then, using the law of total variance, we get that $\text{Var}(S^2) > \text{Var}(\bar{X})$. Based on what I've read, the two above theorems imply that, if we define, say, a sufficient statistic $T_1(\mathbf{X})$ and a complete sufficient statistic $T_2(\mathbf{X})$ for some parameter $\varphi$, then, under some circumstances, we can say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$. However, I'm having trouble understanding this last part. I've read over various notes on the subject, but I'm still not sure I understand what it's saying. Why can we say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$? And what are these 'circumstances' that make this inequality valid?

I am currently trying to learn the two related concepts of the Rao-Blackwell theorem and the Lehmann-Scheffé theorem.

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. My understanding is that, using Lehmann-Scheffé, we get that $E[S^2 \mid \bar{X}] = \bar{X}$. Then, using the law of total variance, we get that $\text{Var}(S^2) > \text{Var}(\bar{X})$. Based on what I've read, the two above theorems imply that, if we define, say, a sufficient statistic $T_1(\mathbf{X})$ and a complete sufficient statistic $T_2(\mathbf{X})$ for some parameter $\varphi$, then, under some conditions, we can say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$. However, I'm having trouble understanding this last part. I've read over various notes on the subject, but I'm still not sure I understand what it's saying. Why can we say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$? And what are these 'conditions' that make this inequality valid?

I am currently trying to learn the two related concepts of the Rao-Blackwell theorem and the Lehmann-Scheffé theorem. My question relates to example 2.16 from this document, and I also found much relevant information in this document.

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. My understanding is that, using Lehmann-Scheffé, we get that $E[S^2 \mid \bar{X}] = \bar{X}$. Based on what I've read, the two above theorems imply that, if we define, say, a sufficient statistic $T_1(\mathbf{X})$ and a complete sufficient statistic $T_2(\mathbf{X})$ for some parameter $\varphi$, then, under some circumstances, we can say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$. However, I'm having trouble understanding this last part. I've read over various notes on the subject, but I'm still not sure I understand what it's saying. Why can we say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$? And what are these 'circumstances' that make this inequality valid?

I am currently trying to learn the two related concepts of the Rao-Blackwell theorem and the Lehmann-Scheffé theorem. My question relates to example 2.16 from this document, and I also found much relevant information in this document.

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. My understanding is that, using Lehmann-Scheffé, we get that $E[S^2 \mid \bar{X}] = \bar{X}$. Then, using the law of total variance, we get that $\text{Var}(S^2) > \text{Var}(\bar{X})$. Based on what I've read, the two above theorems imply that, if we define, say, a sufficient statistic $T_1(\mathbf{X})$ and a complete sufficient statistic $T_2(\mathbf{X})$ for some parameter $\varphi$, then, under some circumstances, we can say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$. However, I'm having trouble understanding this last part. I've read over various notes on the subject, but I'm still not sure I understand what it's saying. Why can we say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$? And what are these 'circumstances' that make this inequality valid?

I am currently trying to learn the two related concepts of the Rao-Blackwell theorem and the Lehmann-Scheffé theorem. My question relates to example 2.16 from this document.

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. My understanding is that, using Lehmann-Scheffé, we get that $E[S^2 \mid \bar{X}] = \bar{X}$. Based on what I've read, the two above theorems imply that, if we define, say, a sufficient statistic $T_1(\mathbf{X})$ and a complete sufficient statistic $T_2(\mathbf{X})$ for some parameter $\varphi$, then, under some circumstances, we can say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$. However, I'm having trouble understanding this last part. I've read over various notes on the subject, but I'm still not sure I understand what it's saying. Why can we say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$? And what are these 'circumstances' that make this inequality valid?

I am currently trying to learn the two related concepts of the Rao-Blackwell theorem and the Lehmann-Scheffé theorem. My question relates to example 2.16 from this document, and I also found much relevant information in this document.

Assume we have the random sample $X_1, \dots, X_n$ with mean $\mu$ and variance $\sigma^2 < \infty$. We have that $E[S^2] = \sigma^2$, where $S^2 = \sum_{i = 1}^n \dfrac{(X_i - \bar{X})^2}{n - 1}$ and $\bar{X} = \sum_{i = 1}^n \dfrac{X_i}{n}$. Now assume the $X_i$ are Poisson random variables with parameter $\lambda$. My understanding is that, using Lehmann-Scheffé, we get that $E[S^2 \mid \bar{X}] = \bar{X}$. Based on what I've read, the two above theorems imply that, if we define, say, a sufficient statistic $T_1(\mathbf{X})$ and a complete sufficient statistic $T_2(\mathbf{X})$ for some parameter $\varphi$, then, under some circumstances, we can say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$. However, I'm having trouble understanding this last part. I've read over various notes on the subject, but I'm still not sure I understand what it's saying. Why can we say that $\text{Var}(T_1(\mathbf{X})) > \text{Var}(T_2(\mathbf{X}))$? And what are these 'circumstances' that make this inequality valid?