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Get the range of the required distribution, in this case, max(X, Y) Find the CDF of this distribution as a function of the known distributions Find the PDF of the distribution by differentiating the CDF
Aug 21, 2011 · M (x) is a function. Taking the maximal number amongst the parameters. max {x1, x2} = {x1, if x1> x2 x2, otherwise. You can define like that the maximum of any finitely many elements. When the parameters are an infinite set of values, then it is implied that one of them is maximal (namely that there is a greatest one, unlike the set {− 1 n ...
$\begingroup$ I prefer $\max\{f(x_1,\ldots,f(x_n)\}$ with curly braces and no parentheses. In this instance, the parentheses don't actually help, and the curly braces remind you that the thing whose maximum is sought is a set rather than a tuple. $\endgroup$
$\begingroup$ I feel like allowing $\arg\max f(x)$ to be either $\in \mathbb{R}$ or $\in \mathcal{P}(\mathbb{R})$ is a very troublesome definition.
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Sep 19, 2017 · One line proof: Since composition of convex functions is convex, we only need to show max (x, y) is convex. But max (x, y) = x + y 2 + | x − y 2 | and | ⋅ | is obviously convex. A function f: Rn → R is convex if and only if its epigraph epif = {(x, t) ∈ Rn × R ∣ f(x) ≤ t} is a convex set.
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But let's take x = 2, then (1 - 2) ^ 2 will be (-1) ^2 which is nothing but 1 and according to op's max function, 1 should be returned. But since you gave the condition of x >= 1, we always return 0 even when x is something like 2. I think in comments what Andre Holzner said is correct.
Rewriting this, E[Z] ≤ logn t + tσ2 2. Now, set t = √2logn σ to get. E[Z] ≤ σ√2logn. The reason Sivaraman set t = \sqrt {2\log {n}}/\sigma is because that is the point at which the upper bound is at a minimum. You can see this by taking the derivative of the bound with respect to t and setting it to zero.
