I have a problem obtaining the Poisson distribution taken at zero.
In class, I saw that the distribution is $$P(Y=y\ | \ Y>0),$$ where $$P(Y=y)=\frac{e^{-\lambda}\lambda^y}{y!},\ \ \ y=0,1,2,...$$ From the definition of conditional probability, I know that $$P(Y=y\ | \ Y>0)=\frac{P(Y=y\ \text{and}\ Y>0)}{P(Y>0)},$$ the denominator is easy to understand, as it's simply $$1-P(Y=0).$$
However, I have trouble understanding how to calculate the numerator. In class, the professor said it was simply $$P(Y=y),$$ but that way, it's not clear to me why the probability that $$Y=0$$ given that $$Y>0$$ is equal to 0. That's my main problem, as I don't understand how to handle this case. Thank you in advance for your explanation.
Think about what the event $$(Y = y) \cap (Y > 0)$$ means. Let's look at an example. Suppose $y = 1$. Then $$(Y = 1) \cap (Y > 0)$$ is equivalent to $Y = 1$. So, as long as $y$ is a positive integer, $\Pr[(Y = y) \cap (Y > 0)] = \Pr[Y = y]$. But what about $y = 0$? Then the event $$(Y = 0) \cap (Y > 0) = \varnothing$$ and $$\Pr[(Y = 0) \cap (Y > 0)] = 0.$$
And here is a simpler example. Let $X$ be a discrete-valued random variable with probability mass function $$\Pr[X = 0] = 0.2, \quad \Pr[X = 1] = 0.1, \quad \Pr[X = 2] = 0.4, \quad \Pr[X = 3] = 0.3.$$
What is $\Pr[X = 1 \mid X > 0]$? It is simply $$\frac{\Pr[(X = 1) \cap (X > 0)]}{\Pr[X > 0]} = \frac{\Pr[X = 1]}{\Pr[X > 0]} = \frac{0.1}{0.8} = \frac{1}{8}.$$
And in general, it is not difficult to see that
$$\Pr[X = x \mid X > 0] = \frac{\Pr[(X = x) \cap (X > 0)]}{\Pr[X > 0]} = \begin{cases}0, & x = 0 \\ 1/8, & x = 1 \\ 1/2, & x = 2 \\ 3/8, & x = 3. \end{cases}$$
The same thing applies to your zero-truncated Poisson distribution.