Expected value of absolute value of the differences, random walk and Brownian motion I want to find the expected value of
\begin{equation}
|\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)|.
\end{equation}
Where $ X_i$ are iid random variables with mean $\mu$ and $B(t)$ is the standard Brownian motion, $ t$ is time and $N$ is a natural number. I know finding the expected value of this function is very rigorous, but I made following inference to calculate it.
since
\begin{equation}
\min_t(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t))\leq |\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)|\leq \max_t(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t))
\end{equation}
and because the $\min$ and $\max$ function will be reached for a t (might be different $t$ for each function), we can calculate the Expected value for that particular function at that point. But for any $t $ that results the  $\min $ or $\max$, the expected value will be zero. Based on this, can we conclude that
\begin{equation}
E[|\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)|]=0.
\end{equation}
I know there is something wrong with this interpretation, however I don't know what. I am also aware of the results related to functional central limit theorem, but I want to find this expected value directly, without further assumptions. If someone can give me any comment on this or the right way to do the calculation of this expectation (or approximating it), I really appreciate it.
 A: 1) Approach when squeeze theorem can be used
If you had the equation:
\begin{align*}
\min_t \mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right)\leq |\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)|\leq \max_t\mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right)\\
\end{align*}
where
\begin{align*}
0= \min_t \mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right)\\
0=\max_t\mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right)
\end{align*}
then
\begin{align*}
0=\min_t \mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right)&\leq |\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)|\leq\\ &\qquad\qquad\max_t\mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right)=0\\
0&=\leq |\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)|\leq 0\\
\end{align*}
Then by the squeeze theorem you are done.

2) Why we can't use squeeze theorem here:
But in fact, what you have is that $t_{min}$ and $t_{max}$ that hold for one particular sequence of $X_i$ values. In general such a $t_{min}$ and $t_{max}$ would not hold over all possible values that each of $\{X_1,\dots,X_n\}$ can take. But the expectation itself is over all possible values of $\{X_1,\dots,X_n\}$ (with the appropriate probability weights). Thus they are not necessarily minimizers of the expectation.
Specifically $\mathbb E[\min_t f(t,X)) \neq \min_t \mathbb E[f(t,X)]$
Thus you can't reach the conclusion you have reached, by the squeeze theorem. (There are of course other ways to reach the conclusion).
Caveat: If for some reason $t_{min}$ and $t_{max}$ will hold for all possible values of $\{X_i\}_{i=1}^N$, then $\mathbb E[\min_t f(t,X)) = \min_t \mathbb E[f(t,X)]$

3. Possible method to show convergence:
We will show that for all fixed t, $\mathbb E\left[\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right]=0$. Fix some t and consider
\begin{align*}
\mathbb E\left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i-\lfloor Nt\rfloor \mu - B(t)\right) 
&= \mathbb E \left(\sum_{i=1}^{\lfloor Nt\rfloor }X_i\right) -\lfloor Nt \rfloor \mu - \mathbb E \left(B(t)\right)\\
&= \sum_{i=1}^{\lfloor Nt\rfloor }\mathbb E X_i -\lfloor Nt\rfloor \mu - 0\\
&=\sum_{i=1}^{\lfloor Nt\rfloor }\mu -\lfloor Nt\rfloor \mu\\
&=\lfloor Nt\rfloor \mu -\lfloor Nt\rfloor \mu\\
&=0
\end{align*}
