Probability distribution function/integration
- Thread starter name14
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You're not very familiar with this sort of thing, are you?
You defined your distribution to be 0 outside the circle. What's the point in integration out there? You can just do [0,r].
It's a circle. You can use integration if you like, but the area will still be \(\displaystyle \pi r^{2}\).
You defined your distribution to be 0 outside the circle. What's the point in integration out there? You can just do [0,r].
It's a circle. You can use integration if you like, but the area will still be \(\displaystyle \pi r^{2}\).
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You can use whatever method you like. The area of the circle will be the same.
If you really want a bivariate uniform distribution on this disk, the height of the distribution will be \(\displaystyle \frac{1}{\pi r^{2}}\) throughout the entire surface and zero (0) on the outside.
If you really want a bivariate uniform distribution on this disk, the height of the distribution will be \(\displaystyle \frac{1}{\pi r^{2}}\) throughout the entire surface and zero (0) on the outside.