Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $t$?

share|cite|improve this question
Are the $V_i$ independent? If so, the answer is easy. – cardinal Oct 4 '11 at 23:15
up vote 1 down vote accepted

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.

share|cite|improve this answer

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 $0 < t < 1$, $$ \begin{align*} P\{X > t\} &< \int_t^1 \int_t^1 \cdots \int_t^1 \beta (1-v_0)^{\beta - 1}\mathrm dv_0 \prod_{i=1}^n \beta v_i^{\beta - 1} \mathrm dv_i\\ &= (1-t)^{\beta}(1 - t^{\beta})^n \end{align*} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.