Assume $x$ is distributed with $F(x)=x^2$ for $0\leq x \leq 1$ and $c\mid x \sim U[0, \lambda \cdot x + 1-\lambda]$ for some $0 < \lambda < 1$. I am trying to find the marginal distribution of c.
The joint density function should be given by $\frac{2x}{\lambda \cdot x + 1-\lambda}$ whenever $x \in [0,1]$ and $c \in [0, \lambda \cdot x + 1- \lambda]$, right?
I tried integrating out $x$ next, but I am unsure about the boundaries. Simply letting $x \geq \frac{c+\lambda -1}{\lambda}$ seems not to be correct, as this lower bound could be negative?
I think should be getting a marginal density for $c$ that integrates to 1 on the support $[0,1]$, regardless of what value $\lambda$ takes, but it never worked no matter what I tried...
First look at the sample space associated with a non-zero joint density:
The $1-\lambda$ location on the figure is for $\lambda=0.3$ but it is meant to be a general y-intercept of upper boundary for $c$.
We see that to construct the value of $Pr[C\leq c_0]$ we need to consider two cases: (1) $c_0 <= 1-\lambda$ and $c_0 > 1-\lambda$.
$$Pr[C\leq c_0 | 0 \leq c_0\leq 1-\lambda \leq 1]=\int _0^1\int _0^{c_0}\frac{2 x}{1-\lambda (1-x)} dc dx = \frac{2 c_0 (\lambda -(\lambda -1) \log (1-\lambda ))}{\lambda ^2}$$
$$Pr[C\leq c_0 | 0 \lt 1-\lambda < c_0 \leq 1]=\\ \int _0^{\frac{c_0+\lambda -1}{\lambda }}\int _0^{1-\lambda (1-x)}\frac{2 x}{1-\lambda (1-x)}dcdx+\int _{\frac{c_0+\lambda -1}{\lambda }}^1\int _0^{c_0}\frac{2 x}{1-\lambda (1-x)}dcdx =\\ \frac{(c_0+\lambda -1)^2-2 c_0 ((\lambda -1) \log (c_0)+c_0-1)}{\lambda ^2}$$
So
$$Pr[C\leq c_0 | 0 \lt 1-\lambda < c_0 \leq 1]=\begin{array}{cc} \{ & \begin{array}{cc} \frac{2 c_0 (\lambda -(\lambda -1) \log (1-\lambda ))}{\lambda ^2} & 0\leq c_0\leq 1-\lambda <1 \\ \frac{(c_0+\lambda -1)^2-2 c_0 ((\lambda -1) \log (c_0)+c_0-1)}{\lambda ^2} & 0<1-\lambda <c_0<1 \\ \end{array} \\ \end{array}$$
Now take the derivative with respect to $c_0$ to obtain the marginal probability density function for $C$:
$$f(c)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{2 (\lambda -(\lambda -1) \log (1-\lambda ))}{\lambda ^2} & 0\leq c_0 \leq 1-\lambda \\ -\frac{2 ((\lambda -1) \log (c_0)+c_0-1)}{\lambda ^2} & 0<1-\lambda< c_0<1 \\ \end{array} \\ \end{array}$$
Here is what the marginal pdf for $C$ looks like for various values of $\lambda$: