Covariance between sum of iid random variables and sum of indicator functions The question is-

Let $X_1,X_2,..,X_n$ be iid random variables from a continuous distribution whose density is symmetric about $0$. Suppose $\mathbb{E}(|X_1|)=2$ and define $Y=\sum_{i=1}^{n}X_i$ and $Z=\sum_{i=1}^{n}I(X_i>0)$. Then calculate covariance between $Y$ and $Z$.

My attempt:
$E(X_i)=0$ for all $i=1(1)n$ because $X$ is symmetric about $0$ and $E(|X|) $ exists. 
Now,
$Cov (Y,Z)=E(YZ)-E(Y)E(Z)$
$=E(YZ)-0$
$=E[(\sum_{i=1}^{n}X_i)(\sum_{i=1}^{n}I(X_i>0)]$
$=(\sum_{i=1}^{n}E[(X_i.I(X_i>0))]$ $+\sum\sum_{i \neq j}E(X_i)E(I(X_j>0)$ as $X_i,X_j$ are independent. 
$=\sum_{i=1}^{n}E[(X_i.I(X_i>0)] +0 $ as $E(X_i)=0$
$ =\sum_{i=1}^{n}\{E[X_i.I(X_i>0)|I(X_i>0)=1]×1/2] + E[X_i.I(X_i>0)|I(X_i>0)=0]×1/2]\}$
$=\sum_{i=1}^{n}E[X_i.I(X_i>0)|I(X_i>0)=1]×1/2] +0$
$=\sum_{i=1}^{n}E[X_i|X_i>0]×1/2]$
$=2n×(1/2)$
$=n$
Is my reasoning correct ? Thanks in advance!
 A: You have a missing part at the end: you didn't show how you get

$\mathbb{E}[ X_i I(X_i > 0) | X_i > 0 ] = \mathbb{E}[|X|]$
  which is something you rely on at the very end.


Below is an argument avoiding conditional expectations altogether.
Assume $X$ is continuous (in particular, no mass at $0$) and symmetric around 0. You have
$$
\mathbb{E}[X \cdot I(X>0)]
= \mathbb{E}[|X| \cdot I(X>0)]
= \mathbb{E}[|X|  - |X|\cdot I(X<0)]
= \mathbb{E}[|X|]  - \mathbb{E}[|X|\cdot I(X<0)] \tag{1}
$$
but, by symmetry of $X$ around 0, $X$ and $-X$ have same distribution, and so $$\mathbb{E}[|X|\cdot I(X<0)] = \mathbb{E}[|-X|\cdot I[-X<0]]  = \mathbb{E}[|X|\cdot I(X>0)]\tag{2}$$
so that, from (1),
$$
\mathbb{E}[X \cdot I(X>0)] = \frac{1}{2}\mathbb{E}[|X|]
$$
allowing you to conclude from what you wrote at the beginning (the first 3 equations).
A: You have correctly shown that
\begin{align}
\mathbb{Cov}\left(\sum_{i=1}^n X_i,\sum_{j=1}^n I(X_j>0)\right)&=\sum_{i=1}^n \sum_{j=1}^n\mathbb{Cov}(X_i,I(X_j>0))
\\&=\sum_{i=1}^n \mathbb{Cov}(X_i,I(X_i>0))+\sum_{i\ne j}^n \underbrace{\mathbb{Cov}(X_i,I(X_j>0))}_{0}
\\&=\sum_{i=1}^n \mathbb{E}(X_1I(X_1>0))
\end{align}
Now just use these equations which follow from the law of total expectation: $$\mathbb{E}(X_1)=\mathbb{E}(X_1 I(X_1>0))+\mathbb{E}(X_1 I(X_1<0))$$
and $$\mathbb{E}(|X_1|)=\mathbb{E}(X_1 I(X_1>0))+\mathbb{E}(-X_1 I(X_1<0))$$
The above can be written using conditional expectations of course but there is no need for that.
