1
$\begingroup$

This is Problem 6.10 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer.

Problem Statement: In a process of sintering (heating) two types of copper powder the density function for $Y_1,$ the volume proportion of solid copper in a sample, is given by $$f_1(y_1)= \begin{cases} 6y_1(1-y_1),&0\le y_1\le 1\\ 0,&\text{elsewhere.} \end{cases} $$ The density function for $Y_2,$ the proportion of type $A$ crystals among the solid copper, is given as $$f_2(y_2)= \begin{cases} 3y_2^2,&0\le y_2\le 1\\ 0,&\text{elsewhere.} \end{cases} $$ The variable $U=Y_1Y_2$ gives the proportion of the sample volume due to type $A$ crystals. If $Y_1$ and $Y_2$ are independent, find the probability density function for $U.$

My Work So Far: Since the variables are independent by assumption, the joint density function is $$f(y_1,y_2)= \begin{cases} 18y_1y_2^2(1-y_1),&0\le y_1,y_2\le 1\\ 0,&\text{elsewhere.} \end{cases} $$ Now then, we have \begin{align*} F_U(u) &=P(U\le u)\\ &=P(Y_1Y_2\le u)\\ &=P(Y_1\le u/Y_2)\\ &=\int_0^1\int_0^{u/y_2}18y_1y_2^2(1-y_1)\,dy_1\,dy_2. \end{align*}

My Question: The problem is that the integral in my last line does not converge, no-how, no-way. I've tried reversing the order of integration, to no avail. Am I missing something basic?

Thanks for your time!

$\endgroup$
2
  • 3
    $\begingroup$ Invariably the solution to problems of this nature is to incorporate the domains explicitly. In other words, the integrand must be multiplied by $\mathscr{I}(0\le y_1,y_2\le 1).$ Equivalently, $u/y_2$ must be replaced by $\min(1,u/y_2).$ As always, it helps to draw a picture of the domain of integration. $\endgroup$
    – whuber
    Aug 26, 2020 at 21:51
  • $\begingroup$ Great, thanks much! $\endgroup$ Aug 26, 2020 at 22:33

1 Answer 1

3
$\begingroup$

From whuber's suggestion, after drawing a picture of the region of integration, the proper limits are: \begin{align*} F_U(u) &=\int_0^1\int_0^{\min(1,\,u/y_2)}18y_1y_2^2(1-y_1)\,dy_1\,dy_2\\ &=\int_u^1\int_0^{u/y_2}18y_1y_2^2(1-y_1)\,dy_1\,dy_2+\int_0^u\int_0^118y_1y_2^2(1-y_1)\,dy_1\,dy_2\\ &=\int_u^1\left(9u^2-\frac{6u^3}{y_2}\right)dy_2+\int_0^u 3y_2^2\,dy_2\\ &=\left(3u^2(3y_2-2u\ln(y_2))\right)\big|_u^1+u^3\\ &=9u^2-3u^2(3u-2u\ln(u))+u^3\\ &=u^2(9-8u+6u\ln(u))\\ f_U(u)&=18u(1-u+u\ln(u)). \end{align*} And of course you have to set up the limits, and build a case structure.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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