Let $X_1, X_2,..X_n $ be Random variables for whether a quantity of certain type of goods were selled at day $n$. In other words, $X_i$ has binary values. Now, define a new random process $Z_n$ that represents the stocks profit at day $n$. $Z_n$ is defined as the following:
$$Z_n = \text{Winning, if $X_n$ and $X_{n-1}$ are true (that quantity was selled on day $n$ and $n-1$)}$$
Otherwise, $Z_n= Loosing$
I want to prove/disapprove that $Z_n$ is memoryless:
If $X_n$ is false, then immediately $Z_n$ becomes $Loosing$. So, there is no dependency on a previous state. Yields, doesn't matter what $Z_{n-1}$ was. In other words, this is a Zeroth-order Markov chain: $$P[Z_n = Loosing] = P[X_n=false| X_n-1=true]P[X_n-1=true] +P[X_n=false| X_n-1=false]P[X_n-1=false] $$ Now, assume that the $X_n$ is true, then we need to look at the previous state $X_{n-1}$, which means that $Z_n$ depends on $X_{n-1}$. Again, assume that we have a $k$ previous $X_{n - j, j = {0,1,..,k}}$ where all of them are true or false. Then, since all $X_i$ are independent, then there are no effects for the previous $X_i$ on the current $X_n$ and that means that $Z_n$ is independent on all previous $Z_{i, i = {0,..,n-1}}$ Which means this process is memoryless.
Does that make sense? I know I'm using plain English to prove this and if my solution is correct, it won't be mathematically rigors.
The process is not memoryless. I will use $1$ and $0$ to denote Winning and Losing.
First, consider $P(Z_{n+1}=1|Z_n=0).$ There are three equally likely scenarios where $Z_n=0$. Either we have $(X_n,X_{n-1})=(1,0),(0,1)$ or $(0,0)$. In the first scenario, where $X_n=1$, we have $P(Z_{n+1}=1)=1/2$, while in the latter two, $X_n=0$, so $P(Z_{n+1}=1)=0$. Therefore, $$P(Z_{n+1}=1|Z_n=0)=\frac13\cdot\frac12+\frac23\cdot0=\frac16.$$
On the other hand, consider $P(Z_{n+1}=1|Z_n=0,Z_{n-1}=0)$. There are now five equally likely possibilities for $(X_n,X_{n-1},X_{n-2})$: $$ 101,100,010,001,000 $$ In the first two, $Z_{n+1}$ occurs half the time, and in the last three, $Z_n$ never occurs, so $$P(Z_{n+1}=1|Z_n=0,Z_{n-1}=0)=\frac25\cdot\frac12=\frac15.$$This is a different answer from before, so the process is not memoryless.