Let $X$ be a random variable and I now assume for simplicity that it is uniformly distributed on $[0,1]$.
Now fix $t \in [0,1]$ and let $Y_t$ be the random variable $Y_t:=\min \{t,X\}$.
I have problems computing for example the density
Computing the cdf is easy:
Of course $P(Y_t \leq y)= 1- P(t>y,X>y)$. Now I tried to split this apart as if $t$ was just random variable independet of $X$ and this gives $P(Y_t \leq y)=y$ if $y<t$ and $1$ if $y\geq t$.
However, I feel that taking the derivative is not the appropriate her since $Y_t$ attains
the value $t$ with positive probability. So $Y_t$ seems to be a random variable that is neither a discrete nor a continuous random variable. So maybe a density does not make sense.
However, in order to be able to compute expectation and so on, I would need a density.
In fact I guess everything becomes easy if I knew on which space $Y_t$ actually lived.
However, I have problems formalizing that. Can anybody help.
Maybe for the uniform distribution everything is easy by direct arguments. But I would like to see a general approach which is valid for $X$ having an continuous distribution.
Best Answer
Your approach is correct. The distribution of $Y_t$ is partly absolutely continuous with respect to the Lebesgue measure, and partly discrete. A formal notation for the distribution $P_{Y_t}$ of $Y_t$ is $$ \text{d}P_{Y_t}(x)=\mathbf{1}_{[0,t]}(x)\text{d}x+(1-t)\delta_t(\text{d}x). $$ But this is only a notation. More to the point, for every Borel subset $B$, $$ P(Y_t\in B)=\text{Leb}(B\cap[0,t])+(1-t)\mathbf{1}_B(t). $$ And to compute expectations, for every bounded measurable function $u$, use $$ E(u(Y_t))=\int_0^tu(x)\text{d}x+(1-t)u(t). $$ In the general case, if $X$ has density $f_X$, one gets $\text{d}P_{Y_t}(x)=f_X(x)\mathbf{1}_{x<t}\text{d}x+P(X\ge t)\delta_t(\text{d}x),$ hence $$ P(Y_t\in B)=P(X\in B\cap(-\infty,t))+P(X\ge t)\mathbf{1}_B(t), $$ and $$ E(u(Y_t))=\int_{-\infty}^tu(x)f_X(x)\text{d}x+u(t)\int_t^{+\infty}f_X(x)\text{d}x. $$