Finding expected area enclosed by the loop I just saw the question the number of loops on lattice? The link is about

number of various non-intersect loops from $(0,0)$ to $(m,n)$ using north and east steps then going back to $(0,0)$ using south and west steps.

and the number of ways are $T_{m,n}=\binom{m+n-2}{m-1}^2-\binom{m+n-2}{m-2}\binom{m+n-2}{n-2}$ for $m\times n$ lattice.
Inspired by the question, I'm thinking of

What is the expected area enclosed by the loop in $m\times n$ lattice?

For example, the area (denoted as $A$) below is $A=42$:

Obviously, $\max(A)=mn$, $\min(A)=m+n-1$.
Any algebraic expression for this?

EDIT 1:
Thanks to Arthur, to be more specific, my question is equivalent to

Consider $m\times n$ lattice. Express number of non-intersect loops with area $A=k$ (between $\max(A)$ and $\min(A)$) in terms of $k$ (or $m, n$).

If there is a function $f_{m,n}(k)$ to express the answer, then I define the expected area be
$$E_{m,n}=\sum_{k=\min(A)}^{\max(A)}P(k)k, \quad\text{where }P(k)=\frac{\text{#(non-intersect loops with $A=k$)}}{\text{#(non-intersect loops)}}=\frac{f_{m,n}(k)}{T_{m,n}}.$$
Example
For $2\times 2$ lattice, there are a total of $3$ possible non-intersect loops:
$2$ non-intersect loops with area $A=3$ and
$1$ non-intersect loop with area $A=4$.
Therefore, $T_{2,2}=3$, $f_{2,2}(3)=2$, $f_{2,2}(4)=1$, and the expected area is
$$E_{2,2}=P(3)\cdot 3+P(4)\cdot 4 = \frac23\cdot 3+\frac 13\cdot4=\frac{10}3.$$
Any algebraic expression for $f_{m,n}$?

EDIT 2:
I'm studying the special case when $m=n$ and $A=2n-1， 2n$.
It's not hard to prove $$f_{n,n}(2n-1)=\binom{2n-2}{n-1}.$$
In the question Proof of a combination identity: $\sum\limits_{i=0}^m\sum\limits_{j=0}^m\binom{i+j}{i}\binom{2m-i-j}{m-i}=\frac {m+1}2\binom{2m+2}{m+1}$, Mike Earnest found a fascinating combinatorial proof for the case $A=2n$ and yielded to $$f_{n,n}(2n)=(2n-3)f_{n-1,n-1}(2n-3)=(2n-3)\binom{2n-4}{n-2}.$$
Some thoughts so far:
It is a possible good way (Mike's way) to solve the case of $A=2n+1, 2n+2,...$.
 A: Putting the question algebraically, we are looking for the number of solutions to
the following set of (in)equalities
$$ \bbox[lightyellow] {  
\eqalign{
  & N_{\,S}\left( {A,m,n} \right) = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}  \cr 
  & \left\{ \matrix{
  0 < y_{\,1}  \le y_{\,2}  \le  \cdots  \le y_{\,m}  = n \hfill \cr 
  0 = x_{\,1}  \le x_{\,2}  \le  \cdots  \le x_{\,m}  = u < n \hfill \cr 
  x_{\,k + 1}  < y_{\,k}  \hfill \cr 
  \sum\limits_{k = 1}^m {y_{\,k} }  - \sum\limits_{k = 1}^m {x_{\,k} }  = A \hfill \cr}  \right. \cr} 
} \tag{1} $$
corresponding to the number of (strictly) non-intersecting loops, from $(0,0)$ to $(m,n)$
defining an area equal to $A$.
The $x$'s and $y$'s are the ordinates of the lower and upper branch respectively.

A first consideration we can make from the geometrical sketch is that, rotating it cw and flipping it vertically,
we obtain another valid loop, with the same area. That means that we have
a symmetry in $n$, $m$.
$$ \bbox[lightyellow] {  
N_{\,S} \left( {A,m,n} \right) = N_{\,S} \left( {A,n,m} \right)
} \tag{2} $$
If we include $u$ among the parameters, so that
$$
N\left( {A,m,n} \right) = \sum\limits_{u = 0}^{n - 1} {N\left( {A,m,n,u} \right)} 
$$
then we can establish for $N(A,m,n,u)$ a recursive relation.
It is not difficult to deduct, either from the sketch and from system (1) that, since for $m=1$
$$
N_{\,S} \left( {A,1,n,u} \right) = \left[ {0 = u < n} \right]\left[ {A = n} \right]
$$
(where $[P]$ denotes the Iverson bracket)
then
$$
\begin{gathered}
  N_{\,S} \left( {A,m,n,u} \right) =  \hfill \\
   = \sum\limits_{\left\{ {\begin{array}{*{20}c}
   {0\, \leqslant \,X\; \leqslant \,A}  \\
   {0\, \leqslant \,x_{\,m - 1} \, \leqslant \,u\, < \,y_{\,m - 1} \; \leqslant \,n}  \\
 \end{array} } \right.} {N_{\,S} \left( {A - X,m - 1,y_{\,m - 1} ,x_{\,m - 1} } \right)\;N_{\,S} \left( {X,1,n - u,0} \right)}  =  \hfill \\
   = \left[ {u < n} \right]\sum\limits_{0\, \leqslant \,x_{\,m - 1} \, \leqslant \,u\, < \,y_{\,m - 1} \; \leqslant \,n}
     {N_{\,S} \left( {A - \left( {n - u} \right),m - 1,y_{\,m - 1} ,x_{\,m - 1} } \right)}  \hfill \\
   = \sum\limits_{0\, \leqslant \,x_{\,m - 1} \, \leqslant \,u\, < \,y_{\,m - 1} \; \leqslant \,n}
 {N_{\,S} \left( {A - \left( {n - u} \right),m - 1,y_{\,m - 1} ,x_{\,m - 1} } \right)} \quad \left| {\;2 \leqslant m} \right. \hfill \\ 
\end{gathered} 
$$
Understanding that $N_{\,S}$ is null for negative values of any of the parameters
and for $m=0$ defining 
$$
N_{\,S} \left( {A,0,n,u} \right) = \left[ {0 = A} \right]\left[ {1 = n} \right]\left[ {0 = u} \right]
$$
we can formulate the recurrence in a way valid for all non-negative values of the parameters
$$ \bbox[lightyellow] {  
\eqalign{
  & N_{\,S} \left( {A,m,n,u} \right)\quad  =   \cr 
  &  = \left[ {0 = A} \right]\left[ {0 = m} \right]\left[ {1 = n} \right]\left[ {0 = u} \right] +   \cr 
  &  + \sum\limits_{0\, \le \,k\, \le \,u\, < \,j\; \le \,n} {N_{\,S} \left( {A - \left( {n - u} \right),m - 1,j,k} \right)}  \cr} 
} \tag{3} $$
