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

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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.
serbskak
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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?
 
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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 '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...
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