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?


$Z$ 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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ I meant possible values; edited. $\endgroup$ – gunes Jan 4 '20 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.