Almost surely convergence of indicator function (pdf of a uniform?)

Question

I'm taking a statistics course and I'm having trouble figuring out how to solve convergence problems. The question is as follows:

1. (Modes of Convergence) Consider Xn(ω) a sequence of rv and X(ω) a rv both defined on the probability space (Ω, F, P), where Ω ≡ [0, 1] denotes the unit interval in the real line, F is the Lebesgue sigma algebra over [0, 1] and P is the Lebesgue measure. Recall the indicator function of a set A ⊆ Ω is 1{A} : Ω → {0, 1}.

   b) Let $X_n$ = $n*I_{(0,1/n)}$ . Does $X_n$ converge to zero almost surely? Does it converge in mean?

My professor has taught the following theorem:

Theorem. Let $(X_n)$ be a sequence of random variables and X a random variable defined on the same probability space. If

$\sum_{n=1}^\infty P(||X_n - X|| > \epsilon) < \infty$

Then $X_n$ tends almost surely to $X$.

I can't seem to proceed on the answer. I believe I'll be left with something divided by $n$ inside the sum so that it sums to less than $\infty$. Also the form of $X_n$ looks a lot like a uniform but I also can't seem to figure out how to use this information.

Thank you in advance

Answer 1

Look at the question again: it asks if $X_n$ converges (in some sense) to zero, so you have to take as limit to test the r.v. $X\equiv0$.

Now try to calculate $P(|X_n-0|>\epsilon)$. Remind that:

• $X_n(\omega)=n \cdot I_{(0,1/n)}(\omega)$, so you could just use this definition, if you know how to.
• It's pretty obvious but you would have to prove it: $X_n \sim \mathcal U(0,1/n)$; this makes easier the other calculations. Maybe after that, the theorem you mentioned could be helpful.

For instance, let's take $\epsilon=0.3$:

Since $X_1\sim \mathcal U (0,1)$, $$P(|X_1-0|< 0.3)=P(-0.3<X_1<0.3)=P(0<X_1<0.3)=\int_0^{0.3} f_{X_1}(x) \,dx=$$ $$=\int_0^{0.3} I_{(0,1)}(x) \,dx=\int_0^{0.3} \,dx=0.3.$$

In the same way, $X_2 \sim \mathcal U (0,0.5)$, so $$P(|X_2-0|<0.3)=\int_0^{0.3} 2 I_{(0,0.5)}(x) \,dx=2\int_0^{0.3}\,dx=0.6;$$ and $P(|X_3-0|<0.3)=3 \cdot 0.3=0.9$.

But from $n=4$ and on things change: Since $X_4\sim \mathcal U (0,0.25)$ (note that now $0.25<0.3$), $$P(|X_4-0|<0.3)=P(-0.3<X_4<0.3)=P(0<X_4<0.3)=\int_0^{0.3} f_{X_4}(x) \,dx=$$ $$=\int_0^{0.3} 4I_{(0,0.25)}(x) \,dx=4\int_0^{0.25} \,dx=4 \cdot 0.25=1.$$ It's easy to prove that in the general case this probability is also 1, for $n\ge 4$. So $P(|X_n-0|>0.3)$ is $0.7$ for $n=1$, $0.4$ for $n=2$, $0.1$ for $n=3$ and $0$ for $n\ge 4$.

Check that you understand these calculations and check that the sum of the theorem you mention converges for $\epsilon=0.3$ (why is that?). Now try to think of an argument for an arbitrary $\epsilon>0$ and you have the proof.