# Given two uniformly distributed independent random variables what should the PDF of multiplication of them? [duplicate]

I am a probability noob and I was solving the following problem, but my answer doesn't match the book's.

The length and width of panels used for interior doors(in inches) are denoted as Xand Y, respectively. Suppose that Xand Yare independent, continuous uniform random variables for 17.75<x<18.25 and 4.75<y<5.25, respectively.determine the probability that thearea of a panel exceeds 90 squared inches.

I thought the area random variable will have a minimum value of 17.75*4.75=84.3125 Maximum value of area will be 18.25*5.25=95.8125

Also if X and Y are uniformly distributed and independent then area should be uniformly distributed too. Every single value that area can take has equal chance. Please correct me here.

Now with this when I integrate area(A) from 90 to 95.8125 for a constant PDF f(A)=1/(95.8125-84.3125), my answer comes out to be 0.5054. But the answer according to the solutions given at the back of book is 0.499. I am not able to understand where I am doing a mistake. Can anyone please help me understand where I am going wrong?

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## marked as duplicate by studiosus, drhab, Norbert, Dan, Alexander Gruber♦Jun 2 '14 at 21:23

A Start: Let random variable $W$ represent the area. The distribution of $W$ is not uniform, though it is not terribly far from being uniform.
If we assume independence (which is not entirely plausible), then the joint density of $X$ and $Y$ is $4$ on the rectangle given by $17.75\le x\le 18.25$ and $4.75\le y\le 5.25$.
Draw the rectangle, and draw carefully the hyperbola $xy=90$. The probability that the area is $\gt 90$ is the integral of $4$ over the part of the rectangle which is "above" the hyperbola. So it is $4$ times the area of a certain region.
In general, the probability that $(X,Y)$ lands in a certain region $D$ that is a subset of the region where our joint density "lives" is $4$ times the area of $D$. This is because the pair $(X,Y)$ has uniform distribution on the rectangle. To see the non-uniformity of $W$, imagine small increases $h$ in length and/or width towards the lower end of the range. This will have a smaller effect on the area of the panel than increases of $h$ in length and/or width towards the upper end of the range. – André Nicolas Jun 2 '14 at 16:58
My comment (the first one) was meant to supply some geometric intuition. Roughly speaking imagine a small square, expand its side by $h$, and a bigger square, expand the side by $h$. There is a larger increase of area in the big square. – André Nicolas Jun 6 '14 at 15:00