Let X and Y be jointly distributed r.v.'s such that
$$ Y |X = x \in Bin(n,x)$$ $$ X\in U(0,1) $$
Compute the characteristic function of Y and Cov(X,Y) (without using the explicit distribution of Y).
I'm completely at a loss here, any guidance on how to proceed would be much appreciated.
As the hint already suggests, you should use the conditional expectation. Let us start with the characteristic function:
$$ \mathbb{E}\big[e^{i \xi Y}\big] = \mathbb{E}_X \bigg[ \mathbb{E}\big[e^{i \xi Y} \big\vert X = x\big] \bigg] $$ For given $x$ the characteristic function is easy to compute by means of the binomial theorem and you will come to the conclusion: $$ \mathbb{E}\big[e^{i \xi Y} \big\vert X = x\big] = (1 +x(e^{i\xi} - 1))^n $$ Now $x$ is distributed uniformly in $[0,1].$ Hence the characterisic function is: $$ \mathbb{E}_X[(1 +x(e^{i\xi} - 1))^n] = \int_0^1 (1 +x(e^{i\xi} - 1))^n \ dx = \frac{1}{n+1}\frac{e^{i\xi(n+1)} - 1}{e^{i \xi}- 1} $$ Now let us pass to the covariance: $$ Cov(X, Y) = \mathbb{E}[(X - \mu_X)(Y - \mu_Y)] $$ where with $\mu$ I indicate the mean. Now we know that $\mu_X = 1/2.$ But we do not know $\mu_Y.$ You could get this value from the characteristic function. But in the spirit of this exercise you could also find it with the usual trick: $$ \mathbb{E}[Y] = \mathbb{E}\big[ \mathbb{E}[Y \vert X = x] \big] = \int_0^1nx \ dx = \frac{n}{2}. $$
Finally one can turn to the covariance. But I think the point should be clear now, so I will not write down the solution. All you need to do is apply the same trick... If you need some additional detail please ask.