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?
$$\begin{align*} \mathbb E[Y_t^d] &= \mathbb E\left[\sum_{k=0}^d \binom{d}{k} (1-2t)^k X^k t^{d-k} \right] \\ &= \sum_{k=0}^d \binom{d}{k} (1-2t)^k \mathbb E[X^k] t^{d-k} \\ &= \sum_{k=0}^d \binom{d}{k} (1-2t)^k t^{d-k} \cdot \frac{1}{k+1} \\ &= \sum_{k=0}^d \frac{d!}{(k+1)!(d-k)!} (1-2t)^k t^{d-k} \\ &= \sum_{k=0}^d \frac{1}{d+1} \cdot \frac{(d+1)!}{(k+1)!((d+1)-(k+1))!} (1-2t)^k t^{d-k} \\ &= \frac{1}{(d+1)(1-2t)} \sum_{k=0}^d \binom{d+1}{k+1} (1-2t)^{k+1} t^{(d+1)-(k+1)} \\ &= \frac{1}{(d+1)(1-2t)} \sum_{k=1}^{d+1} \binom{d+1}{k} (1-2t)^k t^{(d+1)-k} \\ &= \frac{1}{(d+1)(1-2t)} \left( -\binom{d+1}{0} (1-2t)^0 t^{(d+1)-0} + ((1-2t) + t)^{d+1} \right) \\ &= \frac{(1-t)^{d+1} - t^{d+1}}{(d+1)(1-2t)}. \end{align*}$$ 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.