Conditional esperance of Poisson variable

Question

Consider a Poisson variable $X$ which has density function : $$f(x)=\frac{\lambda ^x}{x!}e^{-\lambda}, \quad \lambda >0,\quad x=0,1,...$$ I would like to compute $E(X\mid X < 1)$. In my course, it is written that : $$E(X\mid X < 1)=0$$ Is it because the variable function $f$ is only define with $x$ positive, so, because $X$ is discrete, the event $X<1$ is $X=0$ so the mean value of $X$ if $0$ ? Is that correct ?

Answer 1

As has already been stated in a comment, that's correct.

$f(x)$ is not a density function, but a probability mass function, precisely because $X$ is, as you write, discrete.

Also, $f$ is not defined only for positive $x$ but for all non-negative integers $x$, that is, including $0$.

But that's just to correct your imprecise diction; the main point is that, as you write, here the event $X\lt1$ is identical to the event $X=0$.