Convergence Almost Surely

1 Statement

\(X_n\) converges almost surely to \(X\) when: \[ P\left(\lim_{n\rightarrow\infty} X_n = X\right) = 1 \]

It's taken over \(\Omega\) the outcome space: \[ P\left(\lim_{n\rightarrow\infty} X_n(\omega) = X\right) = 1 \]

In other words, the set \(A \subset \Omega\) where there is no convergence has measure 0. (See probability space note).

Andrew Charles
- Notes on how I intuit this example: Think about a sequence \(X_n\) as an experiment. In almost sure convergence, I know that every experiment I run will eventually converge. For convergence in probability, I know that for any margin \(\epsilon\), and for large enough \(n\), I can get an arbitrarily high fraction of my experiments to lie within that margin, but I can't say anything for certain about any given experiment.
Stack Exchange

Let the outcome space be \(\Omega = [0,1]\)
Say we have \(Y_1\), \(Y_{2,1}\), \(Y_{2,2}\), \(Y_{4,1}\), \(Y_{4,2}\), \(Y_{4,3}\), \(Y_{4,4}\), \(Y_{2^i, 1},\cdots,Y_{2^i,2^i}\) etc.
where
- \(Y_1 = 1_{[0,1]}\)
- \(Y_{2,1} = 1_{[0,\frac{1}{2}]}\)
- \(Y_{2,2} = 1_{[\frac{1}{2}, 1]}\)
- etc
You can visualize the distribution of each \(Y_{2^i, j}\) as being 1 at an interval of length \(\frac{1}{2^i}\) at position \(\frac{j}{2^i}\). So, the intervals get smaller and move left to right across \([0,1]\).
Then, the \(Y_{i,j}\) 's converges in probability to \(Y=0\), because the probability that \(Y_{2^i,j} = 1\) is \(\frac{1}{2^i}\), which goes to 0 as \(i\) goes to infinity
But the \(Y_{i,j}\) 's do not converge almost surely to \(Y=0\), because there are no \(\omega\) 's in the outcome space for which \(\lim_{n\rightarrow\infty} Y_n(\omega) = 0\). The sequence of \(Y_{i,j}\) 's will return infinitely often to label \(\omega\) as 1.