Let's say I have two independent random variables $X$ & $Y$ that are uniformly distributed over $0$ and $1$, with pdf:

$$f_X(x) = \begin{cases}1 & 0<x<1 \\ 0 & \text{ otherwise } \end{cases}$$

$$f_Y(y) = \begin{cases}1 & 0<y<1 \\ 0 & \text{ otherwise } \end{cases}$$

These two random variables are related to random variable Z, according to the function:


Then, I'm given a general formula to apply that will solve for the PDF of Z, if given the pdf of X and Y:

$$f_Z(z) = \int \limits_{-\infty}^{\infty} \bigg|\frac{1}{x}\bigg|~f_{XY}\Big(x,\frac{x}{z}\Big)~dx$$

$$f_Z(z) = \int \limits_{-\infty}^{\infty} \bigg|\frac{1}{x}\bigg|~f_{X}(x)~ f_{Y}\Big(\frac{x}{z}\Big)~dx$$

when I apply this formula, I get:

$$f_Z(z) = \int \limits_{0}^{1} \bigg|\frac{1}{x}\bigg|~(1)(1)~dx$$

$$f_Z(z) = \bigg[\ln x\bigg]_{0}^{1} = (0 - \infty) = -\infty$$

When the textbook applies this formula they get:

$$f_Z(z) = \int \limits_{z}^{1} \bigg|\frac{1}{x}\bigg|~(1)(1)\Big)~dx$$

$$f_Z(z) = \bigg[\ln x\bigg]_{z}^{1} = \ln 1 - \ln z$$

$$f_Z(z) = -\ln z$$

I'm wondering, how did they get the y in the lower limited of the integral?

random variables.

  • $\begingroup$ seems like there are two regions of integration? First you need to find point where y=1 and y=x/z intersect, then draw a vertical line through this point.... integrate square to left of this line and integrate curve starting at this point until x=1? Or, would you just integrate only to the right of the line and ignore the square of the left... $\endgroup$ – pico Apr 9 at 14:38

We have for a $z\in(0,1)$ (well, $(0,1)$ is the range of $Z$, we need information only on this interval): $$ \begin{aligned} f_Z(z) &= \int_{-\infty}^{+\infty} \left|\frac{1}{x}\right|\; f_{(X,Y)}\left(x,\color{red}{\frac zx}\right)\; dx \\ &= \int_{-\infty}^{+\infty} \left|\frac{1}{x}\right|\; f_X(x)\; f_Y\left(\color{red}{\frac zx}\right)\; dx \\ &= \int_0^1 \frac{1}{x}\; (1)\; f_Y\left(\color{red}{\frac zx}\right)\; dx \\ &= \int_z^1 \frac{1}{x}\; (1)\; (1)\; dx \\ &=\Big[\ \ln x\ \Big]_{x=z}^{x=1} \\ &=-\ln z\ . \end{aligned} $$ The limits of integration were changed according to $0\le x\le 1$, needed for $f_X(x)\ne 0$, and according to $0\le \frac zx\le 1$, needed for $f_Y(z/x)\ne 0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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