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I am currently studying martingales and I am working on the following problem:

Let $Omega = mathbb N^*$ and associated probability measure
$$forall, k in mathbb N^*, P({{k}})=frac{1}{k}-frac{1}{k+1}$$
For $,n ge 1$, let the $X_n=(n+1),mathbf{1}_{[n+1,, infty[},,$ and $mathscr F_n=sigma({1},{2},dots, {n},[n+1,, infty[)$

Show that:

1. 
   $(X_n)_{n ge 0}$ is a martingale with respect to $mathscr{(F_n)_{n ge 0}}$
   


2. The $(X_n)_{n ge 0}$ sequence converges $a.s.$ but not in $mathbb{L}^1(Omega)$. Is it uniformly integrable?

1. Martingale

I would like to prove that $mathbb{E}[X_nmidmathscr{F_{n-1}}]=X_{n-1}$

First of all every $X_n$ is $mathbb{L}^{infty}(dP)$ hence $mathbb{L}^1(dP)$, and is $mathscr{F}_n$-measurable.

We note that we can write $F_{n-1}$ as a partition consisting of the $n$ elements ${{1},{2},dots, {n-1},[n,, infty[}$.
We can therefore use the formula for the conditional expectation with respect to a σ-algebra generated by that partition.

$$begin{align*} mathbb{E}[X_nmidmathscr{F_{n-1}}] &= mathbb{E}big[mathbf{1}_{[n,infty]}cdot X_n mid [n,infty]big]+sum_{k=1}^{n-1}mathbb{E}big[mathbf{1}_{{k}}cdot X_n mid {k}big]
\ &= mathbb{E}big[mathbf{1}_{[n,infty[}cdot (n+1), mathbf{1}_{[n+1,infty[} mid [n,infty[big] + sum_{k=1}^{n-1}mathbb{E}big[mathbf{1}_{{k}}cdot (n+1), mathbf{1}_{[n+1,infty[} mid {k}big]
\ &= (n+1),mathbf{1}_{[n,infty[}cdot frac{mathbb{E}big[mathbf{1}_{[n+1,infty[}, mathbf{1}_{[n,infty[}big]} {mathbb{P}([n,infty[)} + (n+1),sum_{k=1}^{n-1} mathbf{1}_{{k}}cdot frac{mathbb{E}big[ mathbf{1}_{[n+1,infty[},mathbf{1}_{{k}} big]}{mathbb{P}({k})}
\ &= (n+1),mathbf{1}_{[n,infty[} cdotfrac{mathbb{P}({[n+1,infty[, cap},{[n,infty[})} {mathbb{P}([n,infty[)} + (n+1),sum_{k=1}^{n-1} mathbf{1}_{{k}}cdot frac{mathbb{P}({[n+1,infty[, cap},{{k}})}{mathbb{P}({{k}})}end{align*}$$
The summation on the right-hand side gives $0$ since ${[n+1,infty[, cap},{{k}}=emptyset, forall k, in [1, n-1]$. So we are left with:
$$begin{align*} mathbb{E}[X_nmidmathscr{F_{n-1}}]&= (n+1),mathbf{1}_{[n,infty[}cdot frac{mathbb{P}({[n+1,infty[,})} {mathbb{P}([n,infty[)}
end{align*}$$
We have the telescopic sum: $$ mathbb{P}({[n,infty[})= sum_{k=n}^{infty}Big(frac{1}{k}-frac{1}{k+1}Big) = frac{1}{n}$$
Hence:
$$begin{align*} mathbb{E}[X_nmidmathscr{F_{n-1}}]&= (n+1),mathbf{1}_{[n,infty[}cdot frac{1}{n+1}cdot n = n , mathbf{1}_{[n,infty[}
\ &= X_{n-1} end{align*}$$

2. 
   

   Convergence

   
   
   
   
   
   • Almost sure convergence
   
   
   
   

My guess is that $X_n xrightarrow[infty]{a.s.}{0}$, so I want to prove that $$forall, epsilon ge 0, lim_{{n}to {infty}} mathbb{P},(sup_{kge n} |X_n| ge epsilon) = lim_{{n}to {infty}} mathbb{P}(sup_{kge n}, (k+1),mathbf{1}_{[k+1, +infty]} ge epsilon)=0$$

$$begin{align*} lim_{{n}to {infty}} mathbb{P}(sup_{kge n}, (k+1),mathbf{1}_{[k+1, +infty]} ge epsilon) &= lim_{{n}to {infty}} mathbb{P}({[n+1, +infty]}) \ &= lim_{{n}to {infty} }frac {1}{n+1} = 0 end{align*}$$

Thus, for every $omega in Omega$, for every large enough $n$, $X_n(omega)=0$. Therefore $(X_n)_{n ge 0}$ converges $a.s.$ to $0$.

• Convergence in $mathbb{L}^1(Omega)$
  

I would like to show that
$$mathbb{E}(|X_n|) xrightarrow[ntoinfty]{}{0} $$

We have:
$$begin{align*} mathbb{E}(|X_n|) &= mathbb{E}big[(n+1),mathbf{1}_{[n+1, +infty]}big] = (n+1),mathbb{E}Big[sum_{k=n+1}^{infty}mathbf{1}_{{k}}Big]
\ &= (n+1),sum_{k=n+1}^{infty}mathbb{E}(mathbf{1}_{{k}}) = (n+1),sum_{k=n+1}^{infty} Big( frac{1}{k}-frac{1}{k+1} Big)
\ &= (n+1)cdot frac{1}{n+1} = 1 neq 0 end{align*} $$
showing that $(X_n)_{n ge 0}$ does not converge in $mathbb{L}^1(Omega)$.

• 
  $(X_n)_{n ge 0}$ is uniformly integrable

According to 1 (item 6) convergence in $mathbb{L}^1$ is equivalent to uniform integrability combined with convergence in probability. Since $(X_n)$ does not converge in $mathbb{L}^1$, either one or both requirements are wrong.
But we know $(X_n)$ converges $a.s.$, which implies convergence in probability. Therefore it must be the uniform integrability requirement which is not met.
We conclude $(X_n)$ is not uniformly integrable.

I have edited my initial solution based on the comments.

Here are some remarks.

1. You write $mathcal F_{n-1}$ as a partition consisting of two elements, but actually we have to write it as a partition consisting of $n$ elements and use the formula for the conditional expectation with respect to a $sigma$-algebra generated by a partition, which you did. By the way, it seems that you do not really need $G_n$, $G_1$ and $G_2$.


2. For the convergence in $mathbb L^1$, there is a mistake in the computation of $mathbb Pleft(left[n+1,+inftyright]right)$, while you did it correctly in the previous step.


3. As a consequence of this mistake, the justification of the uniform integrability is not correct (the expectation of the absolute value is one). The fact that there is no convergence in $mathbb L^1$ combined with item 6 of the linked answer gives a justification.

1. You forgot to point out that every $X_n$ is $L^{infty}(dP)$ hence $L^1(dP)$, and is $mathscr{F}_n$-measurable.

I am also a bit puzzled over your « partition formula » in your computation of the conditional expectation. Can you find some reference or more basic proof or explain?

2. Almost sure convergence

I am not sure your criterion is enough: you are basically claiming that if $(Y_n)$ is a decreasing sequence of nonnegative rv, then if $Y_n$ converges to $0$ in probability it converges to $0$ almost surely.

I would suggest the following proof: for every $omega in Omega$, for every large enough $n$ $X_n(omega)=0$. Thus $X_n$ converges as to $0$.

2. 
   $L^1$ convergence: the expected value of every $X_n$ is $1$, (you forgot to take the indicatir function inside $X_n$ into account). You could have noticed the mistake by seeing that the expected value of terms of a martingale has to be constant.

However you cannot have $L^1$ convergence of any subsequence to any random variable $Y$ because then $Y$ is non-negative with expectation $1$; however since $X_n$ converges to $0$ in law, $Y$ must be $0$, a contradiction.

(Note that from Markov inequality $L^1$ convergence implies convergence in probability implies convergence in law).

Uniform integrability: note that over $mathbb{N^*}$ and any probability, a subspace of $L^1$ is compact iff it is bounded and uniformly integrable. Since no subsequence of the $X_n$ converges in $L^1$, and the sequence is bounded in $L^1$, no infinite subset of them is uniformly integrable.