I have a Poisson process of $X_t$ with rate $\lambda=\tilde\lambda.$ Now, given $M_t = X_t + \gamma\cdot t$ I want to uderstand, for which values of $\gamma$, $M_t$ is a martingale.
I know, that in order for $M_t$ to be a martingale, the following should hold (assuming the natural filtration, and given that $X_t$ is the only term of $M_t$ from which the variability comes):
$$ \mathbb{E}\left[M_{n+1}| \mathcal{F_n}\right] = M_n $$
I have applied the above condition on what I have for $M_t$, but I may have done a mistake, since for some reason $\gamma$ for me does not come to be a constant, what seems quite weird..
$$ \mathbb{E}[M_{n+1}|\mathcal{F}_n] = \mathbb{E}[X_{n+1} + \gamma(n+1)|\mathcal{F}_n] = (\text{since }n \text{ is fixed}) = \gamma(n+1) + \mathbb{E}[X_{n+1}|\mathcal{F}_n] =\\= (\text{here I decided to add and subtract }X_n)=X_n + \gamma n + \gamma + \mathbb{E}[X_{n+1}|\mathcal{F}_n] - X_n = M_n +\\+\left(\gamma - X_n + \mathbb{E}[X_{n+1}|\mathcal{F}_n]\right)\implies \mathbb{E}[X_{n+1}|\mathcal{F}_n] + (\gamma - X_n) = 0~(\text{for }M_n\text{ to be a martingale}). $$ That is, we have to have: $\mathbb{E}[X_{n+1}|\mathcal{F}_n] = X_n - \gamma \Longleftrightarrow \mathbb{E}[X_{n+1}] = X_n - \gamma \Longleftrightarrow \tilde\lambda=X_n-\gamma$. So that I finally obtain that: $$ \gamma = X_n - \tilde\lambda. $$ I am totally sure this result is incorrect, since $\gamma$ is definitely not supposed to involve the variable term. But how can I then obtain it, and what is my mistake?
I would appreciate any help, thank you in advance!
In a martingale $(M_t)$ the mean $EM_t=EM_{t+1}$. So we must have $E(X_t+\gamma t)=E(X_{t+1}+\gamma ((t+1)))$ which gives $t \overline{ \lambda}+\gamma t=(t+1) \overline{ \lambda}+\gamma (t+1)$. Hence, we must have $\gamma =-\overline{ \lambda}$. I will let you try to prove (using independence of increments of $(X_t)$) that this value of $\gamma$ works. (I can post details if necessary).