Consider the Bernoulli probability mass function $$f(x) = p^x (1 - p)^{1-x}; x = 0,1$$ Find the corresponding tilted mass function $f_t(x)$. Is the tilted mass function a Bernoulli mass function?
Attempted solution - From my understanding the tilted mass function is the moment generating function. So, the tilted mass function is given by \begin{align*} f_t(x) = \mathbb{E}[e^{tx}] &= \sum_{x} e^{tx}f(x)\\ &= e^{t(0)}f(0) + e^{t(1)}f(1)\\ &= (1-p) + e^{t}p\\ &= 1 - p + e^{t}p \end{align*}
Attempted solution part 2 - The tilted density of $f$ is given by $$f_t(x) = \frac{e^{tx}f(x)}{M(t)}$$ where $M(t)$ is the moment generating function. In our case the moment generating function of the Bernoulli probability mass function is $$M(t) = \mathbb{E}[e^{tx}] = \sum_{x} e^{tx}f(x) = 1 - p +e^{t}p$$ Thus we have $$f_t(x) = \frac{e^{tx}p^{x}(1-p)^{1-x} }{1-p + e^{t}}$$ How do I check to see if this tilted mass function is a Bernoulli mass function? Any suggestions are greatly appreciated.
If you write everything as an exponential family, your life will be easier.
The Bernoulli's original mass function is \begin{align} f(x) &\propto \exp\left( x \log\left(\frac{p}{1-p}\right) \right) \end{align}
The tilted version has mass function: \begin{align} f_t(x) &\propto \exp(tx) f(x) \\ &= \exp\left( x \left(\log\left(\frac{p}{1-p}\right) + t\right) \right), \end{align} which is clearly still Bernoulli. It has natural parameter $\ell \triangleq \log\left(\frac{p}{1-p}\right) + t$, which corresponds to probability $$\frac{e^\ell}{1+e^\ell}. $$
Your original mass function is
\begin{align} f(x) &= p^x (1-p)^{1-x} \\ &\propto p^x (1-p)^{-x} \\ &= e^{x \log p - x \log(1-p)} \\ &= e^{x \log\left(\frac{p}{1-p}\right)} \end{align}