If \\(X\\) and \\(Y\\) are independent real random variables with densities \\(f\\) and \\(g\\) then \\(X+Y\\) has density\u00a0 \\(f*g\\). This density is bounded as soon as \\(f\\) or \\(g\\) is bounded since \\[\\left\\Vert f*g\\right\\Vert_\\infty\\leq \\min(\\left\\Vert f\\right\\Vert_\\infty,\\left\\Vert g\\right\\Vert_\\infty).\\] One may ask if \\(XY\\) has similarly a bounded density. The answer is\u00a0 unfortunately negative in general. To see it, we note first that when \\(X\\) and \\(Y\\) are non negative then \\(XY\\) has density \\[t\\in\\mathbb{R}_+\\mapsto \\int_0^\\infty\\!\\frac{f(x)}{x}g\\left(\\frac{t}{x}\\right) dx.\\] Now if\u00a0 for instance \\(X\\) and \\(Y\\) are uniform on \\([0,1]\\) then \\(XY\\) has density \\[t\\mapsto -\\log(t)\\mathbf{1}_{[0,1]}(t)\\] which is unbounded... If \\(X\\) is\u00a0 non negative with density \\(f\\) then \\(X^2\\) has density \\[t\\in\\mathbb{R}_+\\mapsto \\frac{f(\\sqrt{t})}{2\\sqrt{t}}.\\] For instance when \\(X\\) is uniform then \\(X^2\\) has (unbounded) density \\[t\\mapsto \\frac{1}{2\\sqrt{t}}\\mathbf{1}_{[0,1]}(t).\\] The density of \\(X^2\\) is bounded if \\(f\\) is bounded and \\(f(t)=O(t)\\) as \\(t\\to0\\) (imposes \\(f(0)=0\\)).<\/p>\n","protected":false},"excerpt":{"rendered":"
If \\(X\\) and \\(Y\\) are independent real random variables with densities \\(f\\) and \\(g\\) then \\(X+Y\\) has density \\(f*g\\). This density is bounded as soon…<\/p>\n