MHB Probability function (p.f) of a random variable

Thread starter serbskak
Start date Nov 3, 2014
Tags

Function Probability Probability function Random Random variable Variable

AI Thread Summary

To find the probability function of the Bernoulli random variable W derived from the Poisson random variable T, the relationship between the two variables is established. The probability that T equals zero is given by P(T=0) = (λ^0/0!)e^(-λ). Consequently, the probability that W equals one is equal to P(W=1) = P(T=0), while the probability that W equals zero is P(W=0) = 1 - P(T=0). Therefore, the probability function for W can be expressed as P(W=1) = e^(-λ) and P(W=0) = 1 - e^(-λ). This provides a clear method to determine the probability function for the random variable W.

Nov 3, 2014

serbskak

If one has a Bernoulli random variable W that is derived from a Variable T (Poisson λ), by the following rules W = (if T=0 then W=1 and if T>0 then W=0), I am having trouble finding the pf for W. Any suggestions about how to proceed forward?

Physics news on Phys.org

Nov 3, 2014

Evgeny.Makarov

Gold Member

MHB

Welcome to the forum.

The distribution for $T$ says that $P(T=0)=\frac{\lambda^0}{0!}e^{-\lambda}$. By assumption $P(W=1)=P(T=0)$ and $P(W=0)=1-P(T=0)$.

Thread 'Value of intuitionistic logic'

I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.

View full post »

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

View full post »