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I am self-learning about DCT and MCT and related Lemma's. I understand the theorems as constructed, but I am struggling to apply them.

As an example:

$X_n=|Z|^{1/n}$ where $Z\sim N(0,1)$

I am trying to do the following:

a)Identify $X$ such that $X_n\rightarrow X$ a.s

b)Determine if $E(X_n)\rightarrow E(X)$ as $n\rightarrow\infty$

I don't think the answer requires exact calculation of expectation, but seeing that done explicitly might help me get my footing.

Work/attempt: This is not a homework question, so any level of detail is extremely appreciated and welcomed.

This is what I know of MCT/DCT:

Monotone Convergence: $f_1,f_2...f_n\uparrow f$ as $n\rightarrow\infty\Rightarrow \lim_{n\rightarrow\infty}\int f_nd\mu=\int fd\mu$

Dominated Convergence: Given measurable functions $f_1,f_2...$ and $|f_n|\le g$ for some $g$, and if $f_n\rightarrow f \mu-a.e.$, then $\lim_{n\rightarrow\infty}\int|f_n-f|d\mu=0$ and $\lim_{n\rightarrow\infty}\int f_nd\mu=\int fd\mu$

So in my example $X_n$ is a series of nth roots of standard normal RVs. I'm not seeing any monotone convergence, so I look to DCT.

Now the trick is to find a function, $g$, that is always greater than or equal to my series. I'm not sure what $X$ this converges to, which makes it hard to find $g$. My first guess is something like $g=|Z|^n$, but I have no theoretical grounding for this choice, and it's not clear to me that it even meets my criteria.


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In the DCT, one tries to get $|f_n|\leqslant g$ with $g$ integrable *and independent of $n$*. Hence $g=|Z|^n$ cannot do anyway. –  Did Jun 22 '12 at 5:45

1 Answer 1

up vote 1 down vote accepted

a) $X=\mathbf 1_{Z\ne0}$ (consider separately the cases $Z(\omega)=0$ and $Z(\omega)\ne0$ and proceed).

b) Since $X_n\leqslant1+|Z|$ for every $n\geqslant1$ (consider separately the cases $|Z(\omega)|\lt1$ and $|Z(\omega)|\geqslant1$ and proceed), Lebesgue dominated convergence theorem implies that $\mathrm E(X_n)\to\mathrm E(X)=\mathrm P(Z\ne0)=1$.

Note: One can also apply Lebesgue monotone convergence theorem twice, separately to the random variables $U_n=|Z|^{1/n}\,\mathbf 1_{|Z|\leqslant1}$ and $V_n=|Z|^{1/n}\,\mathbf 1_{|Z|\gt1}$. To see this, note that $X_n=U_n+V_n$ for every $n\geqslant1$, the sequence $(U_n)$ is nondecreasing, the sequence $(V_n)$ is nonincreasing, $U_n\to U=\mathbf 1_{0\lt|Z|\leqslant1}$ and $V_n\to V=\mathbf 1_{|Z|\gt1}$. Hence $\mathrm E(X_n)\to\mathrm E(U)+\mathrm E(V)=1$.

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