construction of Martingales Consider a random sample of independent and identically distributed random variables with mean 1
. Consider another random variable which is the product of the first n of such random variables as these. Show that    the second random variable forms a Martingale.
To answer this question I considered that the expectation of the product is the product of expectations and hence it is one which is a real number.
Now the the expectation of the future random variables given all the past is the expectation of the present given the present which is just the present. 
Is that proof okay 
 A: We have the "sample" $X_1, X_2, \dots$
Now define $Y_n := X_1 \cdot X_2 \cdots \cdot X_n$
Let $\{ \mathcal F_n\}$ be generated by $\{X_n \}_{n\geq 1}$. In particular that makes $Y_n$ $\mathcal F_n$-adapted. Then we have $E[Y_{n+1} \mid \mathcal F_n]  = E[Y_{n}\cdot X_{n+1} \mid \mathcal F_n]$. Since $Y_n$ is $\mathcal F_n$-measurable, 
$$
E[Y_{n}\cdot X_{n+1} \mid \mathcal F_n] = Y_{n}\,E[X_{n+1} \mid \mathcal F_n] = Y_n\,E[X_{n+1}]=Y_n
$$
A: Just to make Slug Pue's answer more clearer 
Let the i.i.d sample be $ \lbrace  X_i \rbrace _{i=1} ^{\infty}$
We construct a new random variable $\displaystyle  Z_n = \prod _{i=1} ^{n} X_i$
First we must show that $ E(Z_n )$ is finite first as follows:
$\displaystyle  E(Z_n )=E( \prod _{i=1} ^{n} X_i)  $
Since the $ X_i$'s are independent then we must have 
$$\displaystyle  E(Z_n )= \prod _{i=1} ^{n}E( X_i) = \prod_{i=1} ^{n}  1 =1^n=1 \in \mathbb{R} $$
This clearly shows that  $ E(Z_n )$ is finite.
Hence, the new random variable(r.v)    $\displaystyle  Z_n = \prod _{i=1} ^{n} X_i$ has satisfied the first axiom of the fundamental (basic) definition of Martingale process.
Now we want to show that $  E[Z _{n+1} \mid Z_n,..,Z_2,Z_1]=Z_n$
$$ E[Z _{n+1} \mid Z_n,..,Z_2,Z_1] =E[ \prod _{i=1} ^{n+1} X_i \mid Z_n,..,Z_2,Z_1] = E[X_{n+1} \prod _{i=1} ^{n}X_i \mid Z_n,..,Z_2,Z_1]$$
Since $X_{n+1}$ is independent of $\prod _{i=1} ^{n}X_i $
we write $$ E[Z _{n+1} \mid Z_n,..,Z_2,Z_1] = E(X_{n+1} \mid Z_n,..,Z_2,Z_1) E(\prod _{i=1} ^{n}X_i \mid Z_n,..,Z_2,Z_1)$$
similarly $X_{n+1}$ is independent of each of the $ Z_n,..,Z_2,Z_1$ and since  $\displaystyle  Z_n = \prod _{i=1} ^{n} X_i$,  we write 
$$ E[Z _{n+1} \mid Z_n,..,Z_2,Z_1] = E(X_{n+1} ) E(\prod _{i=1} ^{n}X_i \mid Z_n,..,Z_2,Z)= 1 E(\prod _{i=1} ^{n}X_i \mid Z_n,..,Z_2,Z)=E(\prod _{i=1} ^{n}X_i \mid Z_n,..,Z_2,Z)= E(Z_n\mid Z_n,..,Z_2,Z)=Z_n $$
Since$\displaystyle  Z_n = \prod _{i=1} ^{n} X_i$ has satisfied both axioms of the definition of Martingale, then the process $\lbrace Z_n, n \geq 1 \rbrace 
$ is a Martingale
