# 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 $t$?

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Are the $V_i$ independent? If so, the answer is easy. –  cardinal Oct 4 '11 at 23:15

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.

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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*}