Given Probability Density Function, find pdf of $2X+1$ Textbook question:
Suppose $X$ is a binomial random variable with $n=4$ and $p=2/3$. What is the pdf pf $2x+1$
Since its binomial we have
P$_x(k) = {4 \choose k}(2/3)^k(1/3)^{4-k}$
Solution:
In short,
P$_{2x+1}(k)$ = P$_x(\frac{k-1}{2}) = {4 \choose \frac{k-1}{2}}(2/3)^{\frac{k-1}{2}}(1/3)^{4-\frac{k-1}{2}}$
I dont really understand how they got this solution. I can tell that they solved $2x+1 = k \Rightarrow x = \frac{k-1}{2}$ But I dont see why they would set $2x+1 = k$. My first attempt was basically to solve for the equation $2P_x(k)+1$. Can someone explain this to me why my attempt would be wrong please and how there solution is solved, thank you.
 A: What is the binomial distributions aim? To find the probability of $k$ successes where a success has a probability of $p$ and that of a failure, $1-p = q$. Thus, the pdf of such a r.v that is binomially distributed is $$F_{x}(k) = P(X =k ) = \binom{n}{k} p^k q^{1-k}$$
Now, take a random variable $Y = 2X+1$. What do we now need? The probability of $k$ successes, in other words, the p.d.f of $Y$. Thus, we have, $$F_{Y}(k) = P(Y = k) = P(2X+1=k) = P(X =\frac{k-1}{2})$$
Hope you can take it from here.
A: 
My first attempt was basically to solve for the equation $2 P_X(k)+1$. Can someone explain this to me why my attempt would be wrong please and how there solution is solved, thank you.

Because it makes no sense.   $\mathsf P_{2X+1}(k)$ is the probability that random variable $2X+1$ equals value $k$.   $2\mathsf P_X(k)+1$ is one plus twice the probability that $X$ equals $k$.    They are clearly not the same thing at all.
$$\begin{align}
\mathsf P_{2X+1}(k) &= \mathsf P(2X+1=k) && \text{by definition} \\[1ex] & = \mathsf P(X=\tfrac{k-1}{2}) && \text{by algebra} \\[1ex] &= \mathsf P_X((k-1)/2) &&\text{by definition} \\[1ex] & =\binom{n}{(k-1)/2} {p}^{(k-1)/2}~{(1-p)}^{(2n+1-k)/2}\mathbf 1_{k\in[1,2n+1]\cap (2\Bbb N+1)} && \text{by substitution}\end{align}$$
Because ${X\sim\mathcal{Bin}(n,p)}\iff \mathsf P_X(j) = \binom n j p^j (1-p)^{n-j}\mathbf 1_{j\in[0,n]\cap \Bbb N} $
