Monte Carlo standard error for a sum Suppose that I want to compute $E[X+Y]$ using Monte Carlo simulation and compute the standard error. (Note: $X,Y$ are not necessarily independent) The standard way to do this is to

*

*Consider the estimator $\frac{1}{n}\sum_{k=1}^n(X_k+Y_k)$ where $X_k$ and $Y_k$ are iid random variables with distributions the same as $X,Y$, respectively

*Compute the variance $Var(X_1+Y_1) = Var(X_1)+Var(Y_1) + 2Cov(X_1,Y_1)$

*Compute the standard error as the square root of $\frac{1}{n}Var(X_1 + Y_1)$
I am wondering why the following approach is incorrect. The reason why I think it's incorrect is because I don't get the same formula as above but I don't understand why:

*

*Since $E[X+Y] = E[X]+E[Y]$, approximate $E[X]$ by $\frac{1}{n}\sum_{k=1}^n X_k$ and $E[Y]$ by $\frac{1}{n}\sum_{k=1}^n Y_k$

*Compute the standard error of each which are $\sqrt{\frac{1}{n}Var(X_1)}$ and $\sqrt{\frac{1}{n}Var(Y_1)}$

*Add the standard errors for $E[X]$ and for $E[Y]$
Clearly, the two approaches give different results. Can anyone help me clarify the issue?
 A: Let $\bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i$ and $\bar{Y}_n = \frac{1}{n}   \sum_{i=1}^n Y_i$ .
If
$$
Z_n := \bar{X}_n + \bar{Y}_n
$$
Then, since $X_i$, $Y_i$ are iid,
\begin{align*}
\text{Var}(Z_n) &= \text{Var}(\bar{X}_n ) + \text{Var}(\bar{Y}_n) \\
&= \frac{1}{n}\text{Var}(X_1 ) + \frac{1}{n}\text{Var}(Y_1 ) 
\end{align*}
Thus the standard error of $Z_n$ is
$$
\sqrt{\text{Var}(Z_n)} =\sqrt{ \frac{1}{n}\text{Var}(X_1 ) + \frac{1}{n}\text{Var}(Y_1 )}
$$
Since $\sqrt{x+y} \leq \sqrt{x} + \sqrt{y}$,
$$
\sqrt{\text{Var}(Z_n)} \leq \sqrt{ \frac{1}{n}\text{Var}(X_1 ) } + \sqrt{\frac{1}{n}\text{Var}(Y_1 )}$$
So in the case of indenpendent variables, the variance of the sum is the sum of the variances, but the standard error of the sum is not the sum of the standard errors.
For example if we take $\text{Var}(X_1 ) = \text{Var}(Y_1 ) = \sigma^2$ we have
$$
\sqrt{\text{Var}(Z_n)} = \sqrt{ \frac{2}{n}\sigma^2} = \sqrt{2} \sqrt{\frac{\sigma^2}{n}}
$$
while
$$
\sqrt{ \frac{1}{n}\text{Var}(X_1 ) } + \sqrt{\frac{1}{n}\text{Var}(Y_1 )} = 2 \sqrt{\frac{\sigma^2}{n}}
$$
