![](\"wp-content\/uploads\/images\/xy.png\")
\nThe distribution of \\(XY\\) can be derived from its CDF \\(F_{XY}(xy \\le k)\\), which is the total shaded area shown. The red curve is the function \\(y=\\frac{k}{x}\\). This area is thus given by:<\/p>\n
Here is a problem recently described to me. Apparently there is a more elegant solution (which may give more insight), but I don’t see it yet.<\/p>\n
The problem: \\(X, Y, Z\\) are independent random variables uniformly distributed over [0,1]. What is the distribution of \\((XY)^Z\\)?
\n
\nThe conventional solution is to find the distribution of \\(XY\\) first, then of \\((XY)^Z\\).<\/p>\n
\nThe distribution of \\(XY\\) can be derived from its CDF \\(F_{XY}(xy \\le k)\\), which is the total shaded area shown. The red curve is the function \\(y=\\frac{k}{x}\\). This area is thus given by:<\/p>\n
\\(\\int_k^1 \\frac{k}{x} dx + k = k \\ln(x) \\vert_k^1 + k = -k \\ln(k) + k\\), for \\(k>0\\).<\/p>\n
The PDF is the derivative of the above:<\/p>\n
\\(f_{XY}(k) = -\\ln(k)\\), for \\(k>0\\). The density is not well defined at \\(k=0\\).<\/p>\n
![](\"wp-content\/uploads\/images\/xyz.png\")
\nThe second part is to find the distribution in question. Here, the red curve is the function \\(z=\\frac{\\ln(k)}{\\ln(xy)}\\). The CDF \\(F_{(XY)^Z}((xy)^z \\le k)\\) is the total area to the left of the red curve. A column slice shaded in blue has probability per unit of \\(z\\) as labeled. Thus, the CDF is:<\/p>\n
\\(\\int_0^k f_{XY}(v) dv (1-\\frac{\\ln(k)}{\\ln(v)}) = \\int_0^k -\\ln(v) dv (1-\\frac{\\ln(k)}{\\ln(v)})\\)
\n\\( = \\int_0^k -\\ln(v) dv + \\int_0^k \\ln(k) dv\\)
\n\\( = -v \\ln(v) + v \\vert_{\\downarrow 0}^k + k \\ln(k) = -k \\ln(k) + k + k \\ln(k) = k\\), for \\(k>0\\).<\/p>\n
For \\(k=0\\), we can fill in 0, because the minimum value of \\((XY)^Z\\) is 0, so it is the minimum of the support of its CDF.<\/p>\n
So amazingly, \\((XY)^Z\\) has the CDF of (and is) a uniform distribution. How does that happen? What’s the intuition?<\/p>\n","protected":false},"excerpt":{"rendered":"
Here is a problem recently described to me. Apparently there is a more elegant solution (which may give more insight), but I don’t see it yet. The problem: are independent random variables uniformly distributed over [0,1]. What is the distribution of ?<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[1155,534,967,1153,1159,1158,1154,5,1156,39],"class_list":["post-137","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-conventional-solution","tag-curve","tag-frac","tag-independent-random-variables","tag-insight","tag-int","tag-ln-x","tag-math-math","tag-pdf","tag-problem"],"_links":{"self":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/posts\/137","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=137"}],"version-history":[{"count":0,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=\/wp\/v2\/posts\/137\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=137"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=137"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.yhuang.org\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=137"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}