Let $X$ and $Y$ be two discrete random variables (both $X$ and $Y$ Poisson law and not necessarily independent) defined over a probability space, denoted by $(\Omega, \mathcal{F},\mathbb {P})$.
I am looking for a pratical definition of $\mathbb{E}(X\mid Y \geq 1)$. So I was wondering,can we define the conditional expectation as follows $$ \mathbb{E}(X\mid Y \geq 1) = \sum_{i=1}^n i \times \mathbb{P}(X=i \mid Y \geq 1) $$ for fixed $n\in \mathbb{N}^*$.
On
${Y \geq 1}$ is an event in ${\mathcal {F}}$ with non-zero probability, and $X$ is a discrete random variable, the conditional expectation of $X$ given ${Y \geq 1}$ can again be written $$ \mathbb{E}[X\;|\, Y \geq 1 ] = \frac{1}{\mathbb{P}[Y \geq 1]} \sum_{i=1}^n \, i\,\mathbb{P}[\{X = i\}\, \cap \, \{Y \geq 1\} ]$$
I am wondering if we can develop $\mathbb{P}[\{X = i\}\, \cap \, \{Y \geq 1\}$ more explicitly?
Intuitive answer. Probability mass functions work exactly the same in the conditional and unconditional cases. For a random variable $X$ following a Poisson law over $n$ tries, you have $$ \mathbb{E}(X)=\sum_{i=1}^ni\cdot \mathbb{P}(X=i), $$ where $\mathbb{P}(X=i)$ is the probability mass function. Conditioning over the event $Y\geq 1$ on both sides gives the expression you write.
In general, conditional expectations/density functions/... have most of the times the same properties as the non-conditional counterparts.
Short answer. As stated in the comments to your question, what you write is correct through the definition of conditional expectation in the discrete case.