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Context

Let $Y\sim \text{Bernoulli}(p)$ with probability mass function $$ \mathbb{P}(Y = y) = p^y(1 - p)^{1-y} $$ Define a new random variable $Z = a + bY$. What is the probability mass function of $Z$?

My Working

According to this set of notes we follow these steps:

  1. Write $Y = \frac{Z - a}{b} =:v(Z)$
  2. Find derivative $\frac{d v(Z)}{dZ} = \frac{1}{b}$
  3. Write pmf $$ \mathbb{P}(Z=z) = \left|\frac{1}{b}\right|p^{v(z)}(1 - p)^{1 - v(z)} = \left|\frac{1}{b}\right|p^{\frac{Z - a}{b}}(1 - p)^{1 - \frac{Z - a}{b}} $$

Is this correct?

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$Z$ has two possibilities: $a, a+b$ with probabilities $1-p,p$ respectively. We can use your formulation to write it compactly, but note that we can only substitute $a,a+b$ into this equation: $$P_Z(z)=p^{{Z-a}\over b}(1-p)^{1-{{Z-a}\over b}}$$

Just as we can only substitute $y\in\{0,1\}$ in the original Bernoulli PMF. For other values, the PMF is $0$ because the probability is $0$. That set of notes assumes continuous variables by the way.

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  • $\begingroup$ I don’t understand what you mean by “it has two probabilities $a$ and $a+b$”. Those are just real numbers? $\endgroup$ – Euler_Salter Jan 4 '20 at 19:49
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    $\begingroup$ I meant possible values; edited. $\endgroup$ – gunes Jan 4 '20 at 19:50

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