Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$f(x)=60x^3(1-x)^2,\quad 0<x<1.$$

Let $U_1,U_2,...$ and $V_1,V_2,...$ i.i.d. random variables with distribution $U(0,1)$. We build a random variable X as follows: $$U_1\leq60V_1^3(1-V_1)^2/60=V_1^3(1-V_1)^2,$$ If thats not true, we try with $U_2$ and $V_2$:

if $$U_2\leq V_2^3(1-V_2)^2,$$ then $X=V_2$.

If not, we try with $U_3$ and $V_3$, etc. Show that X has density $f(x)$.

It's from an old probability test and a have no idea how to start, could use some hints

share|cite|improve this question
1 ? – VSJ Nov 26 '12 at 5:36

$\def\P{\mathbb P}$So let's first find the index used in defining $X$, that is define $$ T(\omega) := \min \left\{k\in \mathbb N \mid U_k(\omega) \le f\bigl(V_k(\omega)\bigr)/60\right\} $$ if this minimum exists $T(\omega) := \infty$ otherwise. Then $T \colon \Omega \to [0,\infty]$ is measurable. We have \begin{align*} p &:= \P(U_k \le f(V_k)/60)\\ &= \int_{[0,1]^2} \chi_{\{x \le f(y)/60\}} \,d(x,y)\\ &= \int_0^1\int_0^1 \chi_{\{[0,f(y)/60]\}}(x)\, dx\, dy\\ &= \int_0^1 \frac 1{60} f(y)\, dy\\ &= \int_0^1 y^3(1-y)^2\, dy\\ &= \frac 1{60} \left[ 10y^6 - 24y^5 + 15y^4\right]_0^1\\ &= \frac 1{60}. \end{align*} We have therefore \begin{align*} \P(T= k) &= \P\bigl(U_1 > f(V_1)/60, \ldots, U_{k-1} > f(V_{k-1})/60), U_k \le f(V_k)/60\bigr)\\ &= \P(U_1 > f(V_1)/60)\cdots \P(U_{k-1} > f(V_{k-1})/60)\P(U_k \le f(V_k)/60)\\ &= (1-p)^{k-1}p \end{align*} and hence $\P(T < \infty) = 1$. By definition, we have $X = U_T$ and hence, for $x \in (0,1)$: \begin{align*} \P(X \le x) &= \P(U_T \le x)\\ &= \sum_{k=1}^\infty \P(U_T \le x, T = k)\\ &= \sum_{k=1}^\infty \P(U_k \le x, T=k)\\ &= \sum_{k=1}^\infty \P(U_1 > f(V_1)/60, \ldots, U_{k-1} > f(V_{k-1})/60, U_k \le \min\{f(V_{k-1})/60, x\})\\ &= \sum_{k=1}^\infty (1-p)^{k-1} \int_0^x f(y)/60 \, dy\\ &= \frac 1{60p}\int_0^x f(y)\, dy\\ &= \int_0^1 f(y)\, dy. \end{align*} That is, $X$ has density $f$.

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.