PMF function: joint pmf question

Question

For the PMF of $X$ $$P(X = x) = (0.5)^x \;\text{ for }\;x\in\mathbb{Z}^+,$$

would the PMF of $Y= X-1$ just be

$$P(Y = x) = (0.5)^{x-1}.$$

Answer 1

$$ \Pr(Y=x) = \Pr(X-1=x) = \Pr(X=x+1) = 0.5^{x+1}. $$

Answer 2

The final random variable can be express as the sum of tow random variables $X$ and $Y$ defined as:

• $P(X = x) = 0.5^x$
• $Y(Y = y) = \delta(y+1)$, where $\delta$ is the Dirac Delta distribution

The final PMF is the inverse Laplace of product of Laplace transform of $X$ and $Y$:

• $PMF = \mathcal{L}^{-1}\{e^s\mathcal{L}\{0.5^x\}\}$.

So the final PMF is $P(x+1)u(t+1)$, where $u(t-a)$ is the Heaviside step function.

• $PMF = 0.5^{x+1}$ for all $x\gt-1$.

Answer 3

if X is a continus random variable whith PMF f that f(x) > 0 for all x, supose that y=H(x) be a function strictly monotonous, so the PMF Y=H(x) is: $$ g(y) = f(x)|\frac{\mathrm{d}x}{\mathrm{d}y}| $$ Solving the question... $$ \\ g(y)=x-1\\ x = y+1\\\frac{\mathrm{d}x}{\mathrm{d}y}=1\\g(y)=0.5^{y+1} $$