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What is the factorisation theorem?
A fundamental theorem in number theory states that every integer n ≥ 2 can be factored into a product of prime powers. This factorisation is unique in the sense that any two such factorisations differ only in the order in which the primes are written.
What is the zero factor theorem?
You use the zero factor theorem to find the results of a quadratic after you have factored it. For example (From the website above): x2+2x−15=0 factored will give (x−3)(x+5)=0 . By the definition of the Zero Factor Theorem, we know that one or both of those factors has to equal zero.
Is factor theorem and remainder theorem same?
Basically, the remainder theorem links the remainder of division by a binomial with the value of a function at a point, while the factor theorem links the factors of a polynomial to its zeros.
Which is a sufficient statistic for λ factorization?
Therefore, the Factorization Theorem tells us that Y = X ¯ is also a sufficient statistic for λ! If you think about it, it makes sense that Y = X ¯ and Y = ∑ i = 1 n X i are both sufficient statistics, because if we know Y = X ¯, we can easily find Y = ∑ i = 1 n X i. And, if we know Y = ∑ i = 1 n X i, we can easily find Y = X ¯.
Is the factorization theorem a proof of sufficiency?
Therefore, using the formal definition of sufficiency as a way of identifying a sufficient statistic for a parameter θ can often be a daunting road to follow. Thankfully, a theorem often referred to as the Factorization Theorem provides an easier alternative! We state it here without proof.
Which is an example of the factorization theorem?
Let’s put the theorem to work on a few examples! Let X 1, X 2, …, X n denote a random sample from a Poisson distribution with parameter λ > 0. Find a sufficient statistic for the parameter λ. Because X 1, X 2, …, X n is a random sample, the joint probability mass function of X 1, X 2, …, X n is, by independence:
Are there more than one sufficient statistic for a parameter θ?
The previous example suggests that there can be more than one sufficient statistic for a parameter θ. In general, if Y is a sufficient statistic for a parameter θ, then every one-to-one function of Y not involving θ is also a sufficient statistic for θ. Let’s take a look at another example.