Proof of Hoeffding lemma I am reading a proof about Hoeffding's lemma

Let $Y$ be a random variable with $E[Y]=0$, taking values in the
bounded interval $[a, b]$ and let $\psi_Y(t)  = \log E[e^{tY}]$. Then $\psi_Y''(t) \leq (b-a)^2/4$.

At some point of the proof it is stated:

Now, let $P$ denote the distribution of $Y$ and let $P_t$ be the
probability distribution with density
$$x \to e^{-\psi_Y (t)}e^{tx}$$ with respect to P. Since, $P_t$ is
concentrated on $[a, b]$, the variance of a random variable $Z$ with
distribution $P_t$ is bounded by $(b-a)^2$/4. Hence, by an elementary
computation,
$$\psi_Y''(t) = e^{-\psi_Y(t)}E[Y^2e^{t Y}]-
 e^{-2\psi_Y(t)}\left(E[Ye^{t Y}]\right)^2= {Var}(Z) \leq (b-a)^2/4.$$

Why $x \to e^{-\psi_Y (t)}e^{tx}$ is the density of the probability distribution $P_t$? Also, why $$\psi_Y''(t) = e^{-\psi_Y(t)}E[Y^2e^{t Y}]-e^{-2\psi_Y(t)}\left(E[Ye^{t Y}]\right)^2= {Var}(Z)$$ and the variable Z is used instead of $Y$? Could you please someone provide some details?
 A: 
Why $x \to e^{-\psi_Y (t)}e^{tx}$ is the density of the probability distribution $P_t$?

They are constructing a new distribution $P_t$ but have defined it in a rather terse way so it is understandable you are confused. The given density is with respect to $P$, so if $Z \sim P_t$ we have
$$E[f(Z)] = \int_a^b f(y) e^{-\psi_Y(t)} e^{ty} \, dP(y) =  e^{-\psi_Y(t)} E[f(Y) e^{tY}] \tag{$*$}$$
for any measurable $f$.
This is known as an exponential tilting of the distribution $P$ of $Y$.

Also, why $$\psi_Y''(t) = e^{-\psi_Y(t)}E[Y^2e^{t Y}]-e^{-2\psi_Y(t)}\left(E[Ye^{t Y}]\right)^2= {Var}(Z)$$ and the variable Z is used instead of $Y$? Could you please someone provide some details?

I'll write $\psi$ in lieu of $\psi_Y$ for brevity.
Taking the derivatives of $\psi$ (and assuming you may push the derivative into the expectation), we have
$$\psi'(t) = \frac{d}{dt} (\log E[e^{tZ}])
= \frac{E[Ye^{tY}]}{E[e^{tY}]} = e^{-\psi(t)} E[Ye^{tY}]$$
and
\begin{align}
\psi'(t) &= \frac{d}{dt}(e^{-\psi(t)} E[Ye^{tY}])
\\
&= e^{-\psi(t)} E[Y^2 e^{tY}] - e^{-\psi(t)} \psi'(t) E[Ye^{tY}]
\\
&= e^{-\psi(t)} E[Y^2 e^{tY}] - e^{-2\psi(t)} (E[Ye^{tY}])^2.
\end{align}
Now consider a random variable $Z \sim P_t$ following the exponentially tilted distribution defined earlier. Applying the definition ($*$), we see that the variance of $Z$ is
$$\text{Var}(Z) = E[Z^2]-  (E[Z])^2
= e^{-\psi(t)} E[Y^2 e^{tY}] - e^{-2\psi(t)} (E[Y e^{tY}])^2.$$
