$\mathbb E[\bar X_n]=0$

Question

A conditional normal rv sequence, does the mean converges in probability, in this question how can i get $\mathbb E[\bar X_n]=0$?

Here is my attempt;

$$\mathbb E[\bar X_n]=\int_{-\infty}^{\infty}\bar x_n f(x)dx$$ $$=\int_{-\infty}^{\infty}\bar x_n\frac{1}{y_k\sqrt{2\pi}}exp[-\frac{1}{2}(\frac{x-y_k}{y_k^2})]dx$$

$$=\frac{1}{y_k\sqrt{2\pi}}\int_{-\infty}^{\infty}\bar x_nexp[-\frac{1}{2}(\frac{x-y_k}{y_k^2})]dx$$

$$=\frac{1}{y_k\sqrt{2\pi}}e^{\frac{1}{2y_k}}\int_{-\infty}^{\infty}\bar x_n e^{-\frac{1}{2}\frac{x_k}{y_k^2}}dx$$

$$=\frac{1}{y_k\sqrt{2\pi}}e^{\frac{1}{2y_k}}\int_{-\infty}^{\infty}\frac{\sum_{i=1}^{n}x_i}{n} e^{-\frac{1}{2}\frac{x_k}{y_k^2}}dx$$ $$=\frac{1}{y_k\sqrt{2\pi}}e^{\frac{1}{2y_k}}\frac{1}{n}\int_{-\infty}^{\infty}\sum_{i=1}^{n}x_i e^{-\frac{1}{2}\frac{x_k}{y_k^2}}dx$$

Answer 1

It's much easier than you think!

Since by hypothesis we know that the conditional mean is $ E[X_i \mid Y_i] = Y_i $ and $Y_i$ is uniform with mean zero, by using the tower property we find:

$$ n\cdot E [ \bar{X}_n ] = E\left[\sum_{i=1}^n X_i \right] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n E[E[X_i \mid Y_i]] = \sum_{i=1}^n E[Y_i] = 0 $$