Bernoulli distribution $Y_i=\exp\{cX_i\}$

Question

Consider the vector $X:=[X_1,...,X_n]'$ of n Bernoulli trials for $i=1,2,..n$. Each Bernoulli trial can take 0 or 1 with probability $P[X_i=1]=\theta \in (0,1)$. Consider the constant $c\in\mathbb{R}$ and the variables $Y_i=exp\{cX_i\}$, for any $i=1,2,...,n$.

I don't get the follow line: Note that $X_i=0,1$, then $Y_i=1,e^c$. We can then equivalently write $Y_i$ as $Y_i=1+(e^c-1)X_i$.

Why can you write $Y_i$ as above? Thanks in advance!

Answer 1

Work out the formula $Y_i = 1 + (e^c-1)X_i$ for the only two cases that matter, namely $X_i\in\{0,1\}$.

If $X_i = 0$, then $1 + (e^c-1)X_i = 1 = e^{c\cdot 0} = e^{cX_i}$ as required.

If $X_i = 1$, then $1 + (e^c-1)X_i = 1 + (e^c-1) = e^c = e^{cX_i}$ as required.