I have a good understanding of basic Bernoulli and Binomial RVs, but this was foundational work in statistics. I am attempting to try and apply my (minimal but increasing) knowledge of measure theory to a tangible concept. I have been working with simple functions, etc. and am trying to utilize only these tools to find expectation:

if $f=\sum_{i=1}^m c_i1_{A_i}$ has distinct, finite c's and disjoint A's, then $\int f du=\sum_{i=1}^m c_i\mu(A_i)$ and if $f$ is measurable and $f_n \uparrow f$ then $\int f du=\lim_{n\rightarrow\infty}\int f_n d\mu$

I want to try and practice (read: learn how to) utilize these ideas on a measure space of an infinite number of Bernoulli trials. I define my space below:


Take $(\Omega,\mathcal{B})=(\{0,1\}^{\infty},\mathcal{B}(\{0,1\}^{\infty}))$ and define an event $\omega\in\Omega$ as $\omega=(x_1,x_2,...)$

Then I defined a probability measure: $P(\{x_1\}$x{0,1}x...$)=\prod_{i=1}^n p^{x_i}(1-p)^{1-x_i}$.

From here I want to find the expectation of RVs such as:

1- $Z(\omega)=\sum_{i=1}^nx_i$

2- $Y=e^{sZ}$ (moment-generating function), and

3- $V(\omega)=\sum_{i=1}^{\infty}r^nx_n$ for positive r.

Using comments below:

$Z(\omega)=\sum_{i=1}^n 1_{A_i}, A_i\subset\Omega, A_i=\{(x_k)_{k\ge 1}\in\Omega|x_i=1\}$

$Y_n(x_1,...,x_n)=exp \left(s\sum_{i=1}^n x_i \right )$

$E_n(Y_n)=\sum_{(x_k)\in\Omega_n}exp \left(s\sum_{i=1}^n x_i \right )\prod_{i=1}^n p^{x_i}(1-p)^{1-x_i}$


$E(V)=\sum_{i=1}^n r^i p, \forall n$

$V(\omega)_n \uparrow V(\omega)\rightarrow E(V(\omega)_n)\uparrow E(V(\omega))$

I showed these to a friend and he had the following comments:

1-Each RV needs to be represented as a simple function, a limit of a nondecreasing sequence of non-negative functions, or a difference of two such limits (whose product is zero).

For example, on $Z$, you need to compute $E(Z)=\sum_{i=1}^n P(A_i)$ and show how each $P(A_i)$ is derived from $P$ as being a unique probability measure satisfying the equality.

Also, if the RV is a limit of simple functions, you have to find the expectation of the simple function in the sequence and take the limit.

Given that I am learning this on my own from scratch, any explicit help would be greatly appreciated. No detail is too much!

  • $\begingroup$ The current answers do not contain enough detail. You might wish to add more detail about this statement. $\endgroup$
    – Did
    Jun 15, 2012 at 19:08
  • $\begingroup$ @did I am brand new to writing proofs and other formal maths.I have not had any formal courses in any of these subjects so for me to learn these concepts I sometimes need every steps shown. I added a decent amount of my thoughts to the question, including some comments from a friend. I hope that helps clarify. $\endgroup$
    – Justin
    Jun 15, 2012 at 19:19

1 Answer 1


The only case requiring measure theory is 3. Note that $Y$, $Z$ and $V_n:\omega\mapsto\sum\limits_{i=1}^nr^ix_i$ could all be defined on the space $(\Omega_n,2^{\Omega_n},\mathrm P_n)$ where $\Omega_n=\{0,1\}^n$ and $\mathrm P_n$ is the probability measure one can guess.

In particular $\mathrm E(V_n)=\mathrm E_n(W_n)$ where $W_n(x_1,\ldots,x_n)=\sum\limits_{i=1}^nr^ix_i$ hence $\mathrm E(V_n)=\sum\limits_{i=1}^nr^ip$ for every $n$. Since $(V_n)_n$ is nondecreasing to $V$, $\mathrm E(V_n)\to \mathrm E(V)$ and you are done.

Edit: To compute $\mathrm E(Y)$ from first principles, note that $\mathrm E(Y)=\mathrm E_n(Y_n)$ where $Y_n$ is defined on $\Omega_n$ by $Y_n(x_1,\ldots,x_n)=\exp\left(s\sum\limits_{i=1}^nx_i\right)$. Hence, $$ \mathrm E_n(Y_n)=\sum\limits_{(x_k)\in\Omega_n}\exp\left(s\sum\limits_{i=1}^nx_i\right)\prod\limits_{i=1}^np^{x_i}(1-p)^{1-x_i}=\left(\sum\limits_{z=0}^1\mathrm e^{sz}p^z(1-p)^{1-z}\right)^n, $$ and $$ \mathrm E(Y)=(1-p+p\mathrm e^s)^n. $$

  • $\begingroup$ I realize that 1 and 2 do not need measure theory, but how can we use the definitions above to find expectation? In order to use the first property I mentioned, we need distinct $c_i$, but how? $\endgroup$
    – Justin
    Jun 8, 2012 at 18:25
  • $\begingroup$ To introduce the requirement that the scalar $a_i$s are distinct and the sets $A_i$s are disjoint is a bad idea. It is much more convenient to define a simple function as any linear combination of indicator functions of any measurable sets. The class of functions one gets is the same and all your problems disappear... $\endgroup$
    – Did
    Jun 8, 2012 at 19:14
  • $\begingroup$ under that framework, the scalars are p and the A's are just $\omega$s? $\endgroup$
    – Justin
    Jun 8, 2012 at 19:38
  • $\begingroup$ No simple function involves p as a scalar here. Which function do you want to write as a simple function? For example, $Z=\sum\limits_{i=1}^n\mathbf 1_{A_i}$ where $A_i\subset\Omega$ is $A_i=\{(x_k)_{k\geqslant1}\in\Omega\mid x_i=1\}$. $\endgroup$
    – Did
    Jun 8, 2012 at 19:42
  • $\begingroup$ I see. What about the MGF then? Is it similar? $\endgroup$
    – Justin
    Jun 8, 2012 at 20:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .