What is the expected number of ones in each columns? I have a random matrix $\mathbf{A}=\left[a_{ij}\right]$ for all $(i,j)\in\{1,\ldots,k\}\times\{1,\ldots,n\}$. Every entry $a_{ij}$ of the matrix $\mathbf{A}$ is generated randomly with exponential distribution. The $a_{ij}$ are i.i.d and have the same parameter $\lambda$.
Now, for each row $i$ of $\mathbf{A}$, I select the argument of the maximum element. That is,
$$x_i=\arg\max\limits_{j} a_{ij}.$$
Let $X_{ij}$ be the binary random variable that is equal $1$ if $x_i=j$, and $0$ otherwise. Also, let $X_j=\sum_{i=1}^nX_{ij}$.
I am interested in calculating the expected number of ones in column $j$. That is,
$$\mathbb{E}\left[X_j\right],$$
for all $j\in\{1,\ldots,n\}$.
How can I solve this problem?
I will give an example to illustrate the problem: Let $n=3$ and $\mathbf{A}$ given by:
$$\mathbf{A}=\begin{bmatrix}
1 & 3 & 6\\
9 & 7 & 10\\
11 & 5 & 8
\end{bmatrix}.$$
Now, given $\mathbf{A}$, I can calculate $\mathbf{X}=[X_{ij}]$ as:
$$\mathbf{A}=\begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 1\\
1 & 0 & 0
\end{bmatrix},$$
since $x_1=3,x_2=3$ and $x_3=1$. Then, I get $X_1=1,X_2=0$ and $X_3=2$.

When I tried to solve the problem I find that 
$$\Pr\left[X_{ij}=1\right]=\dfrac{1}{n}.$$
After my work, I find that
$$\begin{align}\mathbb{E}\left[X_j\right]&=\sum_{j=1}^n\mathbb{E}\left[X_{ij}\right]\\&=1,\end{align}$$
What is weird is that, in my calculation, I never used the fact that the $a_{ij}$ are exponential random variables. Also, why the expected number of ones is $1$?
 A: Since $\mathbf{A}_{k\times n}$ is a k by n matrix ,through your way of picking $X_{ij}$ :
$$\Pr\left[X_{ij}=1| j _ {fixed}\right]=\dfrac{1}{n}$$
And there will be k one's in your Matrix $$\mathbf{X}=[X_{ij}]$$
So, 
\begin{align}\mathbb{E}\left[X_j\right]&=\sum_{j=1}^n\mathbb{E}\left[X_{ij}\right]\\&=\frac{k}{n},\end{align}
It has nothing to do with  how you generate the random matrix A.
A: Symetry arguments are nice (!). Here, $X_{ij}=1$ for row-maximums. Assuming an i.i.d. continuous source distribution, there is a single row-maximum per row, for a total of $k$ row-maximums for the whole matrix; i.e.
$$
\sum_{i,j}X_{ij}=\sum_{i}X_i=k \rightarrow \sum_{i}E(X_i)=k.
$$
As symetry tells us that $E(X_i)$ does not depend on $i$,
$$
k=\sum_{i}E(X_i)=n\,E(X_i)\rightarrow E(X_i)=\frac kn.
$$
BTW, as each element in a column has a probability $1/n$ of being a row-maximum and that these are independent between rows, the $X_i$'s each follow a binomial distribution with parameters $k$ and $1/n$. This is another way of arriving at $E(X_i)=k\times \frac 1n$. Of course, the $X_i$'s are not independent of one another.
