Warner Bros. Discovery
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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
$\begingroup$ 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.
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 ...
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.
Apr 6, 2012 · Since the perimeter is fixed and we know the perimeter, $28 = 2 * height + 2 * width$ any time you increase the height or the width, you must decrease the other. Also, if you maximize either one, then you would have one of them equal to 14 feet, but that forces the other to be 0 feet (so the total perimeter stays 28ft).$\endgroup$. – user23784.
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.
Now although it is not obvious, but using absolute value is also equivalant to using an if statement e.g. defining Max(a,b) = a if a>b else b; Besides using limits, is there another way of finding the maximum or minimum of two numbers?
Oct 19, 2016 · $\begingroup$ I've got the number max{a,b} and a,b∈R max{a,b}=a, a≥b or max{a,b}=b, a<b We can see that max{a,b}≤c only if a≤c and b≤c Now, if a,b,c,d ∈ R , prove that max{a+b,c+d} ≤ max{a,c} + max{b,d} I am really stuck here, I need this for my University. $\endgroup$ –
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