# Change of metric for probability density vs for probability

When one changes the variable in a probability density function, one must account for the jacobian to ensure the elementary probability is constant (eg Derivation of change of variables of a probability density function?): $$p_Y(y) |dy| = p_X(x) |dx| \Rightarrow p_Y(y) = p_X(x) |\frac{dx}{dy}|$$ So for instance if $$X \sim U(0,1)$$ and $$Y=X^2$$, then $$Y$$ too will be between 0 and 1 with pdf: $$p_Y(y) = \frac{1}{2\sqrt{y}}$$

In the case of a discrete distribution, I don't think the jacobian appears since probabilities, and not probability densities, are in play. I would therefore write for discrete probabilities: $$P_Y(y) = P_X(x)$$ So for instance if $$K$$ follows a Poisson distribution, $$P_K(k;\lambda)=\frac{e^{-\lambda}\lambda^k}{k!} \text{ for } k = 0, 1, 2, 3...$$ and if $$N=K^2$$, then: $$P_N(n;\lambda) = \frac{e^{-\lambda}\lambda^{\sqrt{n}}}{\sqrt{n}!} \text{ for } n = 0, 1, 4, 9...$$ Is my reasoning correct?

You are correct: for a transformation $$T$$ we have $$P(T(X) = a) = P(X \in T^{-1}(\{a\})) = \sum_{x \in T^{-1}(a)} f_X(x).$$ If $$T$$ is invertible then $$P(T(X)=a) = f_X(T^{-1}(a)) = (f_X\circ T^{-1})(a)$$ so we are indeed just plugging $$T^{-1}(a)$$ into the density of $$X$$.
With discrete distributions we fundamentally measure the size of a set by just counting how many elements are involved, and then probabilities come from giving a weight to each element of this set. The size of a set is unchanged by a bijection so if $$T$$ is a bijection then the size of $$A$$ is the same as that of $$T(A)$$. More formally, if $$c$$ is the counting measure then $$c = c \circ T^{-1}$$, i.e. $$c$$ is unchanged by pushing forward with $$T$$.