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!

  • 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 '20 at 21:51
  • $\begingroup$ Great, thanks much! $\endgroup$ – Adrian Keister Aug 26 '20 at 22:33

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.


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