2
$\begingroup$

Suppose we have a sequence $X_1, X_2,...$ of non-negative, real random variables (not necessarily increasing) in $L^1$ which converge in probability to an integrable, non-negative random variable $X \in L^1$. Moreover, let's assume

\begin{equation} E(X_n) \rightarrow E(X). \end{equation}

Since the sequence converges in probability, there exists a subsequence converging to $X$ a.s.. This, together with $E(X_n) \rightarrow E(X)$, implies that this subsequence converges in $L^1$ to $X$.

But how can one prove that $X_n \rightarrow X$ in $L^1$?

$\endgroup$
1
  • 1
    $\begingroup$ Show that every subsequence of $(X_n)$ has a further subsequence that converges to $X$ in $L_1$. $\endgroup$ – David Mitra Jul 20 '16 at 20:05
2
$\begingroup$

It seems that you know how to handle the case where the convergence in probability is replaced by almost sure convergence.

Let's do the general case. As David Mitra suggests, the key point is to extract an almost everywhere convergent subsequence.

Suppose that we do not have the convergence in $\mathbb L^1$. Then there exists a positive $\delta$ and an increasing sequence of integers $\left(n_j\right)_{j\geqslant 1}$ such that for any $j\geqslant 1$, $$\lVert X_{n_j}-X\rVert_1\gt \delta.$$ Now define $Y_j :=X_{n_j}$. We have for any $j\geqslant 1$, $$\lVert Y_j-X\rVert_1\gt \delta.$$ Moreover, the sequence $\left( Y_j\right)_{j\geqslant 1}$ converges to $X$ in probability, hence we can find a subsequence $\left( Y_{j_l} \right)_{l\geqslant 1}$ which converges almost surely to $X$. The fact that $$\lVert Y_{j_l} -X\rVert_1\gt \delta$$ for any $l$ together with the case mentioned in the opening post gives a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.