Probability & Statistics Ch.2 Section 3 Question 12

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let $X \sim \text{Poisson}(\lambda)$ . Let $Y = X - 7$ . What is the probability function of Y?

Solution

From Example 2.3.6 in the textbook, we know:

p_X(x) = P(X=x) = \frac{\lambda^x}{x!}e^{-\lambda}

We want the probability of X for $y + 7$ (since Y = X - 7, we have X = Y + 7):

\begin{aligned} p_Y(y) &= P(X = y + 7) \\ &= \frac{\lambda^{y+7}}{(y+7)!} \cdot e^{-\lambda} \end{aligned}

Note: Y can take values -7, -6, -5, …, corresponding to X = 0, 1, 2, …