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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
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$\begingroup$ 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. $\endgroup$
21 de ago. de 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 ...
11 de mai. de 2020 · 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.
19 de set. de 2017 · 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.
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.
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arg min arg min (or arg max arg max) return the input (s) for which the output is minimum (or maximum). For example: The graph illustrates the function f(x) = 2 sin(x − 0.5) + cos(x)2 f (x) = 2 sin (x − 0.5) + cos (x) 2. The global minimum of f(x) f (x) is min(f(x)) ≈ −2 min (f (x)) ≈ − 2, while arg min f(x) ≈ 4.9 arg min f (x ...
26 de out. de 2016 · You can see it in this way. Determinant is the product of all eigenvalues of the Hessian matrix (2 eigenvalues, in the case of two variables). Then checking the sign of determinant is sufficient to tell the sign of eigenvalues, which is a more general way to test the min/max points. FYI: wiki.
