If the random variables $(v_k)$ are i.i.d, are the $(v_k-m_k)$ i.i.d. as well? This is my first time asking, so if my question is lacking, please let me know so I can edit it.

$v_{k}$ is iid across $k$ where $k=1,....K$ and I can represent its CDF as $F(v)$.
(Note : I have very large $K$ for sure)
I am creating a pseudo reservation value such that $\hat{v}_{k} = v_{k} - \max(0, \max_{k^{'}\neq k}(v_{k^{'}}-c_{k^{'}}))$ where $c_{k^{'}}$ is just constant unique to $k^{'}$.
Let's say $\hat{v}_{k}$ follows new CDF called $\hat{F}(\hat{v_{k}})$.
Goal  : I want to show $\hat{F}(\hat{v_k})$ is iid across $k$ ( or at least for majority of $k$)

Attempt is the following.
Assume $v_{1}-c_{1} > v_{2}-c_{2} > v_{3}-c_{c} > ... > v_{K}-c_{K}$. Then, we have the follwing.
$$\hat{v}_{1} = v_{1} - \max(0, v_{2}-c_{2}) \quad k=1$$
$$\hat{v}_{k} = v_{k} - \max(0, v_{1}-c_{1}) \quad \text{for } k = 2,...,K$$
Current Conclusion :
(1) $Pr(\hat{v}_{k}\leq x) = Pr(v_{k} \leq x+ \max(0, v_{1}-c_{1}))$ for $k=2,...K$. Since I can treat $\max(0, v_{1}-c_{1})$ as some exogenous constant term, and $F(v_{k})$ are identically distributed, $\hat{F}(\hat{v_k})$ is identically distributed for at least $k=2,...,K$.
(2) For sets $A_{2},...,A_{K}$ of possible values of $v_{k}$, $k=2,...,K$,
$$Pr(\hat{v}_{2} \in A_{2} \text{ & } ... \text{ & } \hat{v}_{K} \in A_{K})$$
$$=Pr(v_{k} \in \{ x+\max(0, v_{1}-c_{1}) \quad   x \in A_{k}\}) \text{   for } k= 2,...,K$$
$$ = \Pi_{k=2}^{K} Pr(v_{k} \in \{ x+\max(0, v_{1}-c_{1}) \quad   x \in A_{k}\}))  $$
because $v_{2}, ... v_{K}$ are independent and we have exogenous constant term
$$=Pr(\hat{v}_{2} \in A_{2}) ... Pr(\hat{v}_{K} \in A_{K})$$
Thus, can I say that $\hat{F}(\hat{v}_{k})$ is iid for all $k$ except one?
 A: You can't expect the $\mathbf{\hat v}_k$ to be i.i.d. in general. Here is a simple counterexample :
Let $K=2$ and consider $v_1,v_2\sim\mathrm{Uniform}([1,2])$ i.i.d., then let $c_1 = 3$ and $c_2 = 1$. Because the $v_i$ are almost surely between $1$ and $2$, you have that $\max\{0,v_1-c_1\} = 0$ and $\max\{0,v_2-c_2\} = v_2-c_2$ almost surely. Hence, with your definition, it follows that
$$\hat v_1 = v_1 - (v_2-c_2)\,\text{ and }\, \hat v_2=v_2 $$
You can see that $\hat v_1$ and $\hat v_2$ are not independent, and are not identically distributed either.
Actually, the $\hat v_k$ are defined in such a way that they are very likely to not be independent. Indeed, because of term $\max(0, \max_{k^{'}\neq k}(v_{k^{'}}-c_{k^{'}}))$ which appears in all of them, the $\hat v_k$ actually depend on all of the $(v_i)_{1\le i \le K}$, meaning that they will correlate with each other most of the time.
With that being said, it is still possible for them to have the same distribution : Denote $m_k:=\max(0, \max_{k{'}\neq k}(v_{k^{'}}-c_{k^{'}}))$. By the law of total probability we have
$$\begin{align}F_{\hat v_k}(x):=\mathbb P(\hat v_k \le x) &= \mathbb P\left(v_{k} - m_k\le x\right)\\
&=\int_{t\in\mathbb R}\mathbb P\left(v_{k} \le x+t\ |\ m_k = t\right)f_{m_k}(t)dt\\
&=\int_{t\in\mathbb R}\mathbb P\left(v_{k} \le x+t\right)f_{m_k}(t)dt\\
&=\int_{t\in\mathbb R}F_{v}( x+t)f_{m_k}(t)dt \end{align} $$
Where $f_{m_k}$ is the density function of $m_k$.
From that calculation, we can see that if the coefficients $c_k$ are chosen such that all the $m_k$ have the same distribution, then the $\hat v_k$ will have the same distribution as well. A simple example of $c_k$ for which that is the case would be to take all the $c_k$ equal to each other.
