Bounds on expected value and distribution of a product of beta random variables Let $V_1,...,V_n$ be random variables distribution according to the Beta distribution with parameters $\mathrm{Beta}(1,\alpha)$.
Define $X_i = V_i \prod_{j=1}^{i-1} (1-V_j)$  for $i=1,...,n$.
Is there a way to upper bound (or maybe even calculate accurately?): $E[X_i]$? Or maybe $P(X_i > t)$ for some $t$?
 A: I'm assuming that you mean for the $V_i$ to be independent of each other. Then we have
$$ E(X_i) = E(V_i) \prod_{j=1}^{i-1} (1-E(V_j)) $$
by that independence and linearity.  $E(V_i) = 1/(1+\alpha)$ and so you get
$$ E(X_i) = {1 \over 1+\alpha} \left( {\alpha \over 1 + \alpha} \right)^i $$
Computing $P(X_i > t)$ seems a bit harder. My first guess is to look at 
$$ \log X_i = (\log V_i) + \sum_{j=1}^{i-1} \log (1-V_j). $$
If $i$ is reasonably large you should be able to get a decent approximation out of the central limit theorem.
A: I will use $\beta$ instead of $\alpha$ so as to make the notation
more in line with often-used conventions.  
Let $V_0$ denote a Beta random variable with parameters $(1, \beta)$
and $\{V_i \colon 1 \leq i \leq n\}$ denote $n$ Beta random variables with
parameters $(\beta, 1)$ with the $n+1$ random variables being independent.
Thus, $V_0$ has probability density function (pdf)
$\beta(1-v_0)^{\beta - 1}\mathbf{1}_{(0,1)}$ while the pdf of
$V_i$, $i > 0$ is $\beta v_i^{\beta - 1}\mathbf{1}_{(0,1)}$.  The
joint pdf is, of course, the product of these $n+1$ pdfs. Then,
as claimed by @cardinal and spelled out in more detail by @Michael Lugo,
if $X = V_0V_1\cdots V_n$, then 
$$
E[X] = E[V_0V_1\cdots V_n] = \prod_{i=0}^n E[V_i]
 = \left( \frac{1}{1 + \beta} \right ) \left( \frac{\beta}{1 + \beta} \right )^n
$$
Turning to $P\{X> t\}$, this is the integral of the joint pdf of
the $V$'s over the region of $(n+1)$-space where each $v_i < 1$ 
and $v_0v_1\cdots v_n > t$.  Note that $v_i > t$ for each $i$.
 Let us express this integral as an
iterated integral.  Given  $v_1, \ldots, v_{n-1} \in (0, 1)$ such that
$v_0v_1\cdots v_{n-1} > t$.  Then, the innermost integral (with respect
to $v_0$) has lower limit $v_0 = t/v_1\cdots v_n$ and upper limit $1$.
Thus
$$
\begin{align*}
P\{X > t\} &= \int \int \cdots \int_{v_0 = t/v_1\cdots v_n}^1
\beta(1-v_0)^{\beta - 1} \mathrm dv_0 
\prod_{i=1}^n \beta v_i^{\beta - 1} \mathrm dv_i\\
&= \int \int \cdots \int \left (1 - \frac{t}{v_1\cdots v_n}\right)^{\beta}
\beta^n (v_1\cdots v_n)^{\beta - 1}\mathrm dv_1 \cdots \mathrm dv_n\\
&= \int \int \cdots \int \beta^n 
\frac{(v_1\cdots v_n - t)^{\beta}}{v_1\cdots v_n}
\mathrm dv_1 \cdots \mathrm dv_n
\end{align*}
$$
It might be worth pursuing this further but I will leave this for the OP.  
Alternatively, the region of
integration is a subset of the $(n+1)$-dimensional hypercube
specified as $t < v_i < 1 \colon 0 \leq i \leq n$ and thus,
for $0 < t < 1$,
$$
\begin{align*}
P\{X > t\} &< \int_t^1 \int_t^1 \cdots \int_t^1 \beta
(1-v_0)^{\beta - 1}\mathrm dv_0 \prod_{i=1}^n \beta v_i^{\beta - 1} \mathrm dv_i\\
&= (1-t)^{\beta}(1 - t^{\beta})^n
\end{align*}
$$
