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4

No. You can multiply a sufficient estimator by anything (like, say, 𝑛/(𝑛−1)) and it will stay sufficient but stop being unbiased


1

The definition of bias is that the expectation of the estimate minus its true value. By definition, a proof that some estimate $\hat Y$ is unbiased must show that $E[\hat Y]$ equals $Y$. However, one can also prove that an estimator is asymptotically unbiased, or that the bias approaches 0 as N (or T, perhaps, for panel data) approaches 0. A good example for ...


4

Amazingly, transforming an unbiased estimator often results in a biased estimator. This is how the sample standard deviation is a biased estimator, despite the sample variance being unbiased. This fact comes from something called Jensen’s inequality. For a concave function $f$, such as a square root: $$ f(\mathbb{E}[X])\ge \mathbb{E}[f(X)] $$ Equality holds ...


0

Given the MA(1) model $x_t=\mu + z_t + z_{t-1}$, you can think of your vector of observations $\mathbf{x}=(x_1,x_2,\dots,x_n)$ as coming from an intercept-only regression model $$ \mathbf{x} = \mathbf{1}\mu+\boldsymbol\epsilon $$ where $\mathbf{1}=(1,1,\dots,1)^T$, $\boldsymbol\epsilon \sim N(\mathbf{0},\mathbf{\Sigma})$ and $\mathbf\Sigma$ is tridiagonal ...


0

It’s an assumption of his model. Look at the previous page, Eq 5.29. He assumes that $x^{(i)}\sim\mathcal N(\mu,\sigma^2)$, i.i.d. Gaussian. By definition, then $E[x^{(i)}]=\mu$.


0

The equality$$\int h(T(x_1,...,x_k)) dF(x_1)\cdots dF(x_k) = \int h^{[n]}(x_1,...,x_k)dF(x_1)\cdots dF(x_k)$$does not hold for any function $h$. For instance, if$$h(x_1,...,x_k)=x_1$$ $$\int h(T(x_1,...,x_k)) dF(x_1)\cdots dF(x_k)=\mathbb E^F[X_{(1)}]$$ while$$\int h^{[n]}(x_1,...,x_k)dF(x_1)\cdots dF(x_k)=\mathbb E^F[X_1]$$ Hence, I presume there must be an ...


0

First note that you can get very good results even without solving (the very complex problem) of feedback loops. Also, weighting your loss with inverse propensity is usually not enough and it can cause stability problems. When using IPS you will also need to deal with IPS overfitting and controlling the variance (i.e. regulaizer that make your new policy don'...


1

In response to: So how do I find $E(x^2_i)$? Assuming there are no issues with the steps in the derivation up to that point, you can use the following standard result relating expectation, variance and 2nd moments $$\mathbb{E}[X^2_i] = \text{Var}(X_i) + \mathbb{E}[X_i]^2$$ I know nothing of the Rayleigh distribution, but according to wikipedia, it has mean ...


1

The equalities are incorrect, you should have $\sum Y_i^2 - n\bar{Y}^2$ on the RHS First focus on the sum: $$ \begin{aligned} \sum_{i=1}^{n} (Y_i^2 - \bar{Y}^2) &= \sum_{i=1}^{n} [Y_i^2 - 2Y_i \bar{Y} + (\bar{Y})^2 ]\\ &= \sum_{i=1}^{n} Y_i^2 - 2 \bar{Y} \sum_{i=1}^{n} Y_i + (\bar{Y})^2 \sum_{i=1}^{n}1 \\ &= \sum_{i=1}^{n} Y_i^2 - 2n(\bar{Y})...


1

First, recognize that other methods of dealing with missing data either introduce bias (depending on the type of "missingness") or mis-estimate the standard errors of estimates. See Table 1.1 of Stef van Buuren's Flexible Imputation of Missing Data. Second, although multiple imputation (MI) might lead to bias in some circumstances, that does not ...


2

The Lehmann-Scheffé Theorem says that any unbiased estimator that is a function of a complete sufficient statistic is the minimum variance unbiased estimator. Here $\sum_i X_i$ is a complete sufficient statistic for $\lambda$ and $\bar X$ is a function of it, and is unbiased for $\lambda$, so it is best unbiased. You might then ask if this really avoids ...


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