Distribution of a random variable that is produced by another uniformly distributed random variable Assume a simple two-step lottery: the reward (denoted by $Y$) is a random variable and realized as follows:
First: random variable $X$ is drawn from uniform distribution by interval $(a,c)$.
Second: After the realization of $X$, a player determines reward ($Y$) stochastically:

*

*with probability $\alpha$: reward is $bX$  ($b$ is constant)


*with probability $1-\alpha$: reward is $X$
Now, what is the distribution of the random reward $Y$?
If the solution is weighted average on both states: i.e. $Y=\alpha bX+(1-\alpha)X=(\alpha b+1-\alpha )X$
then we can say $Y$ is a linear rescaling transformation of $X$ and so $Y$ has the same distribution as $X$ has. But I am suspicious that $Y$ really is a weighted average of $X$ and $bX$

Let $E$ denote the event that the reward is taken to be $bX$.
Then:$$P(Y\leq y)=P(E)P(Y\leq y\mid E)+P(E^c)P(Y\leq y\mid E^c)=$$$$\alpha P(bX\leq y\mid E)+(1-\alpha)P(X\leq y\mid E)$$
If moreover $1_E$ and $X$ are independent then this reduces to:$$P(Y\leq y)=\alpha P(bX\leq y)+(1-\alpha)P(X\leq y)\tag1$$
Can you take it from here?

Looking at it as weighted average as in your question the stochastic character of the choice is disregarded.
Actually we are dealing with:$$Y=1_EbX+(1-1_E)X\tag2$$where $1_E$ is a random variable (Bernoulli distribution with parameter $\alpha$).
This instead of $\alpha bX+(1-\alpha)X$.
