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As the title says, I'm looking for the marginal densities of $$f (x,y) = c \sqrt{1 - x^2 - y^2}, x^2 + y^2 \leq 1.$$

So far I have found $c$ to be $\frac{3}{2 \pi}$. I figured that out through converting $f(x,y)$ into polar coordinates and integrating over $drd\theta$, which is why I'm stuck on the marginal densities portion. I know that $f_x(x) = \int_{-\infty}^\infty f(x,y)dy$, but I'm not sure how to solve that without getting a big messy integral, and I know the answer isn't supposed to be a big messy integral. Is it possible to instead find $F(x,y)$, and then take $\frac{dF}{dx}$ to find $f_x(x)$? That seems like the intuitive way to do it but I can't seem to find anything in my textbook that states those relationships, so I didn't want to make the wrong assumptions.

whuber
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Jarrod
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1 Answers1

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Geometry helps here. The graph of $f$ is a spherical dome of unit radius. (It follows immediately that its volume is half that of a unit sphere, $(4 \pi /3)/2$, whence $c=3/(2 \pi)$.) The marginal densities are given by areas of vertical cross-sections through this sphere. Obviously each cross-section is a semicircle: to obtain the marginal density, find its radius as a function of the remaining variable and use the formula for the area of a circle. Normalizing the resulting univariate function to have unit area turns it into a density.

whuber
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  • Ahh, it's sort of coming back to me from multivariable calculus. I remember doing problems like that. How do I find the radius as a function of the remaining variable? It still seems like I'm going to have some sort of monster integral left over. – Jarrod Oct 11 '10 at 22:13
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    Let the remaining variable be $y$. Then $x^2 \le 1-y^2$ describes the region over which you have to integrate. Evidently the radius equals $\sqrt{1 - y^2}$, whence the cross-sectional area equals $\pi (1 - y^2)/2.$ That's a pretty simple formula :-). (Remember, the theme here is geometry, not calculus...) – whuber Oct 11 '10 at 22:29
  • Oh, right. That crossed my mind, but it seemed too simple. I guess I was determined for it to be complicated. Thanks! – Jarrod Oct 11 '10 at 22:40
  • I forgot to ask: how does c figure into this? – Jarrod Oct 11 '10 at 22:49
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    In my opinion, Whuber's answer deserves to be upvoted for two reasons. First it answers the question asked, second as a model for how we could in the future handle (explicitly stated) homework questions: this type of answers actually contributes to the learning process and could be a better policy with respect to homework question than that adopted at MO/SO. – user603 Oct 12 '10 at 08:50
  • @Jarrod: you can use c to determine the normalization constant for the marginal density. Alternatively, forget c, derive the functional form of the marginal (that is, up to a constant factor), then integrate the marginal to find its normalizing constant. – whuber Oct 12 '10 at 12:45
  • @Kwak: Thanks for pointing out there is a policy concerning homework questions! I continued this discussion on meta at http://meta.stats.stackexchange.com/q/12/919 . – whuber Oct 12 '10 at 13:12
  • In case anyone was dying to know if there was a way to solve this with calculus, you can instead recognize that $x = \sqrt{1-y^2}$ is of the form $u = \sqrt{a-b^2}$, which has the special integral $\int udx = \frac{1}{2}(xu + a^2 arcsin \frac{x}{a})$, which gives you the same result as wheber's method when integrated from $-\sqrt{1-y^2}$ to $\sqrt{1-y^2}$. My professor discussed this in class today, but I prefer wheber's method for obvious reasons. – Jarrod Oct 12 '10 at 20:19
  • @Jarrod: Thank you for the follow-up. If the purpose of your course is to learn applications of calculus, then your professor's solution has pedagogical merit. If it's to learn about probability, then seeing multiple solutions--including some relatively simple ones--can help you identify the key *probability* ideas and learn them better. (The calculus stuff might then just be a distraction...) – whuber Oct 12 '10 at 21:36