Finding the conditional distribution of the uniform distribution

Question

Let $U$ denote a random variable uniformly distributed over $(0,1)$. Compute the conditional distribution of $U$ given that $$(a)\,\,\,U>a\\ (b)\,\,U<a$$ where $0<a<1.$

For $(a)$ I tried the following, $$f_{U\mid U>a}(u\mid u>a)=\frac{f(u,u>a)}{f_{U>a}(u>a)}=\frac{{\displaystyle \frac{1}{1-a}}}{{\displaystyle \int_a^1\frac{du}{1-a}}}=\frac1{1-a}$$

and for $(b)$ $$f_{U\mid U<a}(u\mid u<a)=\frac{f(u,u<a)}{f_{U<a}(u<a)}=\frac{{\displaystyle \frac{1}{a}}}{{\displaystyle \int_0^a\frac{du}{a}}}=\frac1{a}$$

would this process be correct? Because I feel as if $f_{U>a}(u>a)$ should be equal to $\int_0^1\frac{du}{1-a}$ because we are integrating with respect to $U$ and not $U>a$ and the interval of definition of $U$ is $(0,1)$, or am I wrong?

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Here I just used the definiton given by the book which is $$f_{X\mid Y}(x\mid y)=\frac{f(x,y)}{f_Y(y)}=\frac{f(x,y)}{\int_{-\infty}^\infty f(x,y)\,dx}$$

Answer 1

The conditions do not change the fact that the distribution is uniform, only the interval for which it is defined. Therefore for an interval of $0\le u\le a$, the density function is $\frac{1}{a}$, while for an interval $a\le u\le 1$, the density function is $\frac{1}{1-a}$. Your analysis is correct.

To clarify your concern about the integral, remember that by imposing the condition, the density function is $0$ outside the limit imposed by the condition.

Answer 2

$\begin{align}f_{U\mid U>a}(u)&=\dfrac{f_U(u)~\mathbf 1_{u>a}}{F_U(a)}\\[1ex]&=\dfrac{\mathbf 1_{0<a<u<1}}{1-a} \end{align}$