# A spoof justifying the erroneous statement that convergence in probability implies almost surely convergence

Today after spending a few unproductive hours and not being able to think anything or do anything, I came up (almost via a dream) with a spoof that looks elegant enough to boggle your mind (at least for a few seconds). I usually don’t like posting stuff which seem to confuse people rather than helping them have fun with mathematics or understand something, but I think it can serve as an example helping some readers to understand where they can go wrong in deriving results related to stuff like convergence of sequences (especially those that are related to the convergence modes we are going to speak about in this post). To present it in a less confusing way, we would directly mention what could have gone wrong leading us to an erroneous result.

We (referred to those who already know the basics about the various modes of convergence of random variables) know that every sequence of random variables that converge in probability to a random variable $X$ has a sub sequence which converges almost surely to $X$.

Now, let $(X_n)$ be a sequence of  random variables which converge in probability to some random variable $X$. Then we know that every sub sequence of $(X_n)$ will converge in probability to $X$ (WHY ? Prove it as an exercise). Thus every sub sequence of $(X_n)$ will have a further sub sequence that converge almost surely to $X$. So, $(X_n)$ is a sequence of random variables every sub sequence of which has a further sub sequence converging almost surely to $X$.

Now, recalling a result from convergence of real numbers we know that a sequence $(x_n)$ of reals converge to some real number $x$ if and only if every sub sequence of $(x_n)$ has a further sub sequence converging to $x$. We may get tempted to use the same conclusion here with random variables. But this is wrong. To be precise, this result doesn’t hold true for the almost sure mode of convergence of random variables.

Instead of justifying why this goes wrong for the almost sure mode of convergence, let us see what would have happened if it was true.

This would have implied that $(X_n)$ converges almost surely to $X$.

So in short, we would have got convergence in probability $\implies$ almost sure convergence and we know that this is wrong. Almost sure convergence is a much stronger condition than convergence in probability.

Exercise : Show that a sequence $(X_n)$ of random variables converge in probability to $X$ if and only if every sub sequence of $(X_n)$ has a further sub sequence that converges almost surely to $X$. (That is, in other words the result we mentioned about convergence of sequences of real numbers hold true for the convergence in probability of random variables).