Does this product converges in $L^1$? Suppose $(X_n)$ is a discrete sequence of independent and equally distributed random variables such that : $E[X_3]=1$ and $P(X_3=\alpha)$ +$P(X_3=\beta)=1$.
Does the product $P_n=\Pi_{k=1}^n X_n$ converges in $L^1$?
my attempt is that : $E[P_n]=(E[X]^n)=1$ so it doesn't converge but I'm not sure
Edit: I forgot to add that $\alpha$ is in $]0,1[$ and $\beta \gt \alpha$
 A: A necessary condition for convergence in $L^1(dP)$ is uniform integrability, i.e. that for every $\epsilon > 0$ there exists $K_\epsilon>0$ such that for all $n$, $\int_{|X_n| \geq K_\epsilon} |X_n| dP<\epsilon$.
So now here, what is that integral? It is $\sum_{k : \beta^k \alpha^{n-k} \geq K_\epsilon} {n \choose k} (\beta p_2)^k (\alpha p_1)^{n-k}$. So the sum starts at $k^*+1:=\left \lceil \frac{\log(K_\epsilon)}{\log(\beta)-\log(\alpha)} \right \rceil$ assuming $n \geq k^*+1$.
From the binomial theorem, if it started at $0$ it would be $(\alpha p_1+\beta p_2)^n=1$ so if $n \geq k^*$ then you have $1-\sum_{k=0}^{k^*} {n \choose k} (\beta p_2)^k (\alpha p_1)^{n-k}$. So the question is, what happens when you send $n \to \infty$ in this sum for fixed $k^*$? In this case eventually the last term dominates and so you are essentially looking at $1-{n \choose k^*} (\alpha p_1)^n \left ( \frac{\beta p_2}{\alpha p_1} \right )^{k^*}$ which goes to $1$ (${n \choose k^*}$ has only polynomial growth which loses to the exponential decay of the other term).
Thus $X_n$ is not uniformly integrable and thus it does not converge in $L^1$.
What is interesting about this result is that the process converges in probability to zero (and from numerical simulations seems to converge a.s. also, but I didn't check this). But the large deviations in the binomial are likely enough to spoil the $L^1$ convergence nonetheless, because you have "just the right amount" of "exponential tilting" through the function $x \mapsto \beta^x \alpha^{n-x}$, a classic situation in large deviation theory.
Actually, in view of that observation, you get an alternate, more abstract, less technical proof: $X_n$ converges in probability to $0$, so if it converges in $L^1$ to something then that something must be $0$, but $\| X_n-0 \|_{L^1}=1$ by the binomial theorem.
Side remark: this process is a martingale, since $E[X_{n+1} \mid X_n]=p_2 \beta X_n + p_1 \alpha X_n=X_n$. Doob's second martingale convergence theorem says that uniform integrability is sufficient for convergence in $L^1$ for martingales. So this check of uniform integrability was especially reasonable in this problem, because if we had gotten the other outcome then we would still know what was happening.
