part-ii/2018-121/

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Part II 2018 Probability and Measure: Paper 2, Section II, J

https://questions.tripos.org/part-ii/2018-121/

[eha1234]

(a)

Holder's inequality

(b)

Construct $n_k$ by taking $n_k$ to be sufficiently large that you have control over $\sum_{k<j} \mathbb{E}(Y_k Y_j)$, for instance, you can inductively define $n_k$ so that $| \mathbb{E}(X_{n_i}X_{n_j}| \leq \frac{1}{2^j(j-1)}$

(b)

$L^1$ convergence: $L^2$ bounded implies UI by Holder's inequality. The rest is bookwork.

$L^2$ convergence: No. Take $X_n = n^{1/2} 1_{{0<w < 1/n}}$. Have $\mathbb{E}(X_n) = 1$ and it converges almost everywhere to 0 (hence converges in probability to 0), but it does not converge in $L^2$.

(d)

Define $Z_n = X_n - \mathbb{E}(X_n)$, bounded in $L^2$. Calculate $||\frac{1}{n}\sum_{n=1}^N Z_n ||_2^2$ to see that it converges in $L^2$ to $0$. Now $\frac{1}{N}\sum X_n = \frac{1}{N} \sum Z_n + \frac{1}{N}\sum \mathbb{E}X_n$ so the condition is that $\frac{1}{N}\sum \mathbb{E}X_n$ must converge