# How to find the Bayesian equivalent of of $\bar{X}-\bar{Y}$

Let $$X_1, \dots, X_n$$ i.i.d. from $$N(\mu, \sigma^2)$$; $$Y_1, \dots, Y_m$$ i.i.d. from $$N(\eta, \tau^2)$$. $$X_1, \dots, X_n$$ are independent of $$Y_1, \dots, Y_m$$. And $$\tau^2$$ and $$\sigma^2$$ are considered known.

I want to show that $$\bar{X}-\bar{Y}$$ is a minimax estimator for estimating $$\Delta=\mu-\eta$$ under the square loss.

I can show that the risk function of $$\bar{X}-\bar{Y}$$ is $$R(\bar{X}-\bar{Y}, \mu-\eta)=\frac{\sigma^2}{n}+\frac{\tau^2}{m}$$

If I can find the Bayes estimator $$\delta_\Lambda$$, so I can get the Bayes risk $$r_\Lambda$$. When the variance terms go to infinity if $$r_\Lambda \rightarrow R(\bar{X}-\bar{Y}, \mu-\eta)$$, then $$\bar{X}-\bar{Y}$$ is a minimax estimator by Theorem 1.12.

But I do not know how to derive the Bayes estimator $$\delta_\Lambda$$ for $$\bar{X}-\bar{Y}$$.

I know how to get it for $$\underline{X}~N(\theta, \sigma^2)$$, which is $$\delta_n=\frac{\frac{n}{\sigma^2}\bar{X}+\frac{\mu^*}{b^2}}{\frac{n}{\sigma^2}+\frac{1}{b^2}}$$ As prior distribution for $$\theta$$, the conjugate normal distribution $$N(\mu^*, b^2)$$.

So, the Bayes risk is $$r_\Lambda = \frac{1}{\frac{n}{\sigma^2}+\frac{1}{b^2}}$$. So, $$r_\Lambda \overset{b\to\infty} \to \frac{\sigma^2}{n} = \underset{\theta}SUP R(\theta, \bar{X})$$

Could anybody teach me how to get the Bayes risk for $$\bar{X}-\bar{Y}$$. Thanks!

{From Theory of Point Estimation, E.L. Lehmann p.343

Theorem 1.12: Suppose that $$\{\Lambda_n\}$$ is a sequence of prior distributions with Bayes risks rn satisfying $$r_\Lambda \le r = \lim_{n\to\infty} r_{\Lambda_n}$$, where $$r_{\Lambda_n}=\int R(\theta, \delta_n)d\Lambda_n(\theta)$$ is a Bayes risk under $$\Lambda_n$$, and that $$\delta$$ is an estimator for which $$\underset{\theta}SUP R(\theta,\delta)=r$$ Then (i) $$\delta$$ is minimax and (ii) the sequence $$\{\Lambda_n\}$$ is least favorable.}

• Could you modify the title? You are looking for the Bayesian equivalent of $\bar X-\bar Y$. Commented Apr 23, 2021 at 5:51
• @Xi'an Yes, I modified the title and added the reference which is from Theory of Point Estimation, E.L. Lehmann p.343. Thank you! Commented Apr 23, 2021 at 13:50

It seems like you already know how to calculate the Bayes estimator for a single normal random variable. Now let $$Z = \bar{X} - \bar{Y}$$. By the property of the independent normal random variables, $$Z \sim N\left(\mu - \eta, \frac{\sigma^2}{n} + \frac{\tau^2}{m}\right)$$. Now simply work with $$Z$$ and apply the formula for the Bayes estimator of a normal normal conjugate family to this. (i.e put the $$\theta = \mu- \eta$$ and the proper $$\sigma^2$$ in the bayes estimator formula you have in the question)
• the way that I am using to derive the Bayes estimator $delta_n$ is to find the proportion of the posterior density, i.e., derived from $e^{-\frac{\sum_{i=1}^{n}(X_i-\theta)^2}{2\sigma^2}}*e^{-\frac{(\theta-\mu)^2}{2\tau}}$. But, m and n may not be the same, I do not know how to plug $\sum_{i=1}^{n}X_i$ and $\sum_{i=1}^{m}Y_i$ into the formula. Could you tell me a little bit more? Thanks! Commented Apr 23, 2021 at 15:13
• $Z$ is a single sample from $N(\mu_z, \sigma^2_z)$, Does that help? Commented Apr 23, 2021 at 21:43