Find the PDF of $Y = X^2 + 3$ where $X \sim Poisson(\lambda)$.

Question

I am following the CDF method to calculate the PDF oF Y. Up to this, I have done:

$$F_Y = P(Y \leq x)$$ $$F_Y = P(X^2+3 \leq x)$$ $$F_Y = P(X \leq \sqrt{x-3)}$$ $$F_Y = e^{-\lambda}.\sum_{k=0}^{\sqrt{x-3}} \frac{\lambda^k}{k!}$$

Now I have to differentiate the CDF to get PDF.

$$f_Y = \frac{\partial F_Y}{\partial x}$$

Now how can I proceed from here? In the continuous case, I can use Leibniz rule but I'm not sure about the discrete case.

Answer 1

The CDF methods you showed is for continuous distribution. With discrete laws the pdf (more precisely the pmf, probability mass function) is not the derivative of the CDF.

With a discrete distribution, it is enough to directly calculate the pmf:

$$\mathbb{P}[Y=y]=\mathbb{P}[X^2+3=y]=\mathbb{P}[X=\sqrt{y-3}]=\frac{e^{-\lambda}\lambda^{ \sqrt{y-3}}}{ (\sqrt{y-3})!}\mathbb{1}_{\{3;4;7;12,...\}}(y)$$