# Tight Probability on Hilbert space

I am considering the following problem. Let $(X_j)$ be i.i.d. $N(0,1)$ random variables and $H$ a Hilbert space with orthonormal basis $(e_j)$. Let $$X:=\sum_j \frac{X_j e_j}{j}$$ And for any $\varepsilon>0$, find a compact $K$ so that $\mathbb{P}(X\in K)>1-\varepsilon$.

I have really no idea what to try or why this should work, so any intuition is helpful.

• For any $x \in H$ write it in coordinate form as $\sum x_i e_i$ where $e_i$ is your basis. Hint: The Hilbert Cube $\{x: x_j \in [-\frac{c}{j},\frac{c}{j}]\}$ is compact for any $c$. If you have not seen this proven before, it follows from using the sequential compactness definition of compactness, a diagonalization argument, and the dominated convergence theorem. – Chris Janjigian Apr 10 '13 at 16:25
• I think I see most of the argument from your hint. One question. If we choose K to be the Hilbert cube as you described, why would we not simply have the probability of $X\in K$ be 1 rather than $>1-\varepsilon$? – Keaton Apr 10 '13 at 16:35
• I think the argument I was thinking of is incorrect actually. The probability that all of the normals are less than any fixed constant is zero. My guess is that something involving the Hilbert cube is how to do this though. The essentialy property is that the width of the interval is square summable, so maybe there needs to be a logarithmic correction to the size of the intervals so that the probability is nonzero. – Chris Janjigian Apr 10 '13 at 16:52

What I gave as a hint was almost right, so let me write out a correction. Fix a constant $c>0$ so that the set $K = \{x: x_j \in [\frac{-c\cdot(\ln(j)^2+1)}{j},\frac{c\cdot(\ln^2(j)+1)}{j}]\}$ is compact (the proof is the same as what I described above: all that matters is that the width is square summable). We would like to show that $P( X \in K)$ can be controlled by $c$. Notice that $$P(X \notin K) = P(\exists j \mbox{ : } |X_j| > c(1 + \ln^2(j) )$$ We can control this with $$\sum_j P(|X_j| > c(1 + \ln^2(j) ) \leq$$ $$\sum_j \frac{2}{\sqrt{2 \pi}}\frac{1}{c(1 + \ln^2(j) )}e^{-c(1 + \ln^2(j) )}$$
Once we notice that $e^{-c\ln^2(j)} = (\frac{1}{j})^{c\ln(j)}$ this sum clearly converges and tends to zero as $c \to \infty$.