Why are the moments of my random variable infested with Fibonacci polynomials? Preamble
I've stumbled upon (a particular variant of) Fibonacci polynomials in my work, defined by
$$
F_k(x) = \begin{cases} \qquad\qquad 1, & k=1 \\ \qquad\qquad 1, & k=2 \\ F_{k-1}(x) + x\!\; F_{k-2}(x), & k > 2 \end{cases}
$$
giving rise to a sequence of polynomials starting with
$$\begin{aligned}
\begin{split}
F_1(x) &= 1 \\
F_2(x) &= 1 \\
F_3(x) &= 1 + x \\
F_4(x) &= 1 + 2x\\
\end{split}
&\qquad\qquad
\begin{split}
F_5(x) &= 1 + 3x + x^2\\
F_6(x) &= 1 + 4x + 3x^2 \\
F_7(x) &= 1 + 5x + 6x^2 + x^3 \\
F_8(x) &= 1 + 6x + 10x^2 + 4x^3
\end{split}
\end{aligned}$$
and so forth. I'm very pleased to have encountered them, but I'd like to know what business they think they have in the moments of a random variable ranging over $[0,1]$ that I'm looking at.
Context
I'm doing work with a random variable $X$ which ranges within $[0,1]$, with moments
$$ \mathbb E [X^d] \,=\, 1\big/(d+1) \,.$$
Using $X$, I define another variable $Y_t$ depending on a real parameter $t \in [0,1]$:
$$ Y_t \;=\; {(1-t) X + t(1-X)} \;=\; {t + (1-2t)X} $$ 
Computing the first few moments of $Y_t$ explicitly, what I found was that they form integer polynomials in $(t^2 - t)$, divided by $(d+1)$. After looking up the sequence of coefficients of the integer polynomials at the OEIS, I encountered A011973, which  allowed me to realise that the polynomials governing these moments could be written as
$$
  \mathbb E[Y_t^d] = \tfrac{1}{d+1} F_{d+1}(t^2 - t), \qquad 0 \leqslant d \leqslant 4.
$$
Question.
This is all very nice, but I only have circumstantial evidence: I don't have any proof that the moments of $Y_t$ have anything to do with Fibonacci polynomials. But the formula $F_{d+1}(t^2 - t)/(d+1)$ recovers the exact lower and upper bounds which I know holds of the moments of $Y_t$, and structures like this rarely come up by accident in my line of work.
What business does a recursively generated polynomial sequence like this have doing in the moments of my random variable?
 A: $$\begin{align*}
\mathbb E[Y_t^d]
&= \mathbb E\left[\sum_{k=0}^d \binom{d}{k} (1-2t)^k X^k t^{d-k} \right] \\
&= \sum_{k=0}^d \binom{d}{k} (1-2t)^k \mathbb E[X^k] t^{d-k} \\
&= \sum_{k=0}^d \binom{d}{k} (1-2t)^k t^{d-k} \cdot \frac{1}{k+1} \\
&= \sum_{k=0}^d \frac{d!}{(k+1)!(d-k)!} (1-2t)^k t^{d-k} \\
&= \sum_{k=0}^d \frac{1}{d+1} \cdot \frac{(d+1)!}{(k+1)!((d+1)-(k+1))!} (1-2t)^k t^{d-k} \\
&= \frac{1}{(d+1)(1-2t)} \sum_{k=0}^d \binom{d+1}{k+1} (1-2t)^{k+1} t^{(d+1)-(k+1)} \\
&= \frac{1}{(d+1)(1-2t)} \sum_{k=1}^{d+1} \binom{d+1}{k} (1-2t)^k t^{(d+1)-k} \\
&= \frac{1}{(d+1)(1-2t)} \left( -\binom{d+1}{0} (1-2t)^0 t^{(d+1)-0} + ((1-2t) + t)^{d+1} \right) \\
&= \frac{(1-t)^{d+1} - t^{d+1}}{(d+1)(1-2t)}.
\end{align*}$$
To show equivalence to your choice of definition of Fibonacci polynomials, you need only establish that the recurrence holds; i.e. let $$G_d(x) = \frac{(1-x)^d - x^d}{1-2x}.$$  We aim to show $$F_d(t^2-t) = G_d(t).$$  If this is the case, then $G$ should obey the recurrence $$G_d(t) = G_{d-1}(t) + (t^2-t)G_{d-2}(t).$$  I leave this as a straightforward algebraic exercise, as well as establishing the base cases.  This completes the proof.
