independent random variable sentence in Hindi
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- However, such processes do not consist of independent random variables : indeed, many purely deterministic, non-random systems can be mixing ).
- Nevertheless, because independent random variables are simpler to work with, this reparametrization can still be useful for proofs about properties of the Dirichlet distribution.
- The theorem here relates to the limiting distributions for the minimum or the maximum of a very large collection of independent random variables from the same distribution.
- On the product space S ^ N, starting with " N " independent random variables with probability distribution \ eta _ 0 and elementary transitions
- If some of the random variables are independent of all of the others, then any cumulant involving two ( or more ) independent random variables is zero.
- The setting is : Let X and Y be independent random variables ( on the reals ), with density functions f _ X and f _ Y.
- Fisher defined his new term variance, as the square of the standard deviation, because of the manner in which variances of independent random variables may be added.
- Since independent random variables are always uncorrelated, the equation above holds in particular when the random variables X _ 1, \ dots, X _ n are independent.
- The probabilistic interpretation of this is that, for a sum of n independent random variables X _ 1, \ ldots, X _ n with common distribution F,
- Let \ mathcal { P } be a finite set of mutually independent random variables and \ mathcal { A } be a finite set of events determined by these variables as before.
- The characteristic function approach is particularly useful in analysis of linear combinations of independent random variables : a classical proof of the Central Limit Theorem uses characteristic functions and L�vy's continuity theorem.
- A converse is Raikov's theorem, which says that if the sum of two independent random variables is Poisson-distributed, then so is each of those two independent random variables.
- A converse is Raikov's theorem, which says that if the sum of two independent random variables is Poisson-distributed, then so is each of those two independent random variables.
- The difference between 0.85185 . . . and 0.85558 . . . seems remarkably small when it is considered that the number of independent random variables that were added was only three.
- Observe that, even if \ mu is close to 1, h ( x _ i ) are no longer independent random variables, which is often a problem in the analysis of randomized algorithms.
- *PM : moment generating function of the sum of independent random variables, id = 9628 new !-- WP guess : moment generating function of the sum of independent random variables-- Status:
- *PM : moment generating function of the sum of independent random variables, id = 9628 new !-- WP guess : moment generating function of the sum of independent random variables-- Status:
- To see how this implies the preservation of log-convexity by independent sums, suppose that " X " and " Y " are independent random variables with log-concave distribution.
- Given a set of bad events \ mathcal { A } we wish to avoid that is determined by a collection of mutually independent random variables \ mathcal { P }, the algorithm proceeds as follows:
- :It also follows that the probability generating function of the difference of two independent random variables " S " = " X " 1 & minus; " X " 2 is
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