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The definition of zero-truncated Poisson (ZTP) distribution is:

\begin{align} g(k;\lambda)= P(X = k | X > 0) &= \frac{f(k;\lambda)}{1-f(0;\lambda)} \\[8pt] &= \frac{\lambda^ke^{-k}}{k!(1-e^{-\lambda})} \\[8pt] &= \frac{\lambda^k}{(e^\lambda - 1)k!} \end{align}

The definition of conditional probability is: $$ P(A|B)=\frac{P(A\cap B)}{P(B)} $$

Now, $A = f(k;\lambda)$ and $B = 1-f(0;\lambda)$, but how is $A\cap B = f(k;\lambda)$?

Source: wikipedia

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Note that A(k) refers to $X=k$, B refers to $X>0$.

So - for all $k>0$ is A(k) is true B is true as well meaning: $ \forall k>0:P(A(k)\cap B) = P(A(k)), P(A(0)\cap B)=0$

Now I see that I can stop here, but I already wrote it so...

So , what you got is $$P(A(k)|B)=\frac{p(A(k))}{P(B)}=\frac{\lambda^k}{(1-P(X=0)) k!}=\frac{\lambda^k}{(1-e^\lambda)k!}=\frac{\lambda^ke^{-\lambda}}{(1-e^{-\lambda})k!}$$

which is actually the real distribution

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  • $\begingroup$ Could you explain how you get across that last equality (i.e. what you did to get from the LHS to the RHS of it?) $\endgroup$ – Glen_b Jul 28 '17 at 2:09

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