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Let $X_1,X_2,...,X_n$ be the iid sample from $N(\theta,1)$.

Then how can I compute the correlation coefficient between sample mean $\bar X$ and $X_1$?

In order to get the $\rho$, I need to take care of $E[(X_1-\theta)(\bar X-\theta)]$, which equals $E(X_1\bar X)-\theta^2$.

To get $E(X_1\bar X)$, i need joint pdf of $\bar X$ and $X_1$, which requires $\rho$.

Where did I stuck?

Please help me with it.

  • 0
    Joint distribution of $(X_1,\overline X)$ is bivariate normal. But you can calculate the covariance directly. $$\operatorname{Cov}\left(X_1,\frac{1}{n}\sum_{i=1}^n X_i\right)=\frac{1}{n}\left[ \operatorname{Var}(X_1)+\sum_{j=2}^n\underbrace{\operatorname{Cov}(X_1,X_j)}_{=0}\right]=\frac{1}{n}$$ So, $$\operatorname{Corr}(X_1,\overline X)=\frac{1}{\sqrt n}$$ – 2018-11-08

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Here is one approach. Let $V = \sum_{i=2}^nX_i$. The distribution of $V$ is clearly still normal, and $V$ and $X_1$ are independent! This means we know the joint distribution.

Now consider the expression: $X_1\bar{X} = X_1(V+X_1)/n$. Therefore:

\begin{equation} E[X_1\bar{X}] = \frac{1}{n}\int_R\int_Rx(v+x)f_{X_1,V}(x,v)dx dv \end{equation}

Update: You can also directly compute $$E(X_1\bar{X}) = E(X_1(V+X_1)/n) = \frac{1}{n}\left[E(X_1)E(V) + E(X_1^2)\right]$$