Show that a random variable $T_x$ is uniformly distributed given that $T$ is uniformly distributed? We have a lifetime $T$, which is uniformly distributed over $(0,b)$. We then introduce a new r.v., $T_x=T-x$, which is defined on $0<x<b$.
We want to show that given $T>x$, the variable $T_x$ is also uniformly distributed over $(0,b-x)$.
My understanding is that we're just shifting this variable. I feel like it's pretty obvious that it would still be uniform, since we're just consistently shifting the value of $T$ by the value of $x$ within the interval. However, I don't know how to go about showing this using notation consistent with probability theory. Any help or hints are appreciated.
 A: I'll try not to give away the entire answer, but as you said it does seem pretty obvious. The issue is then converting this into rigorous mathematics! Let's proceed as follows.
We're interested in $P(T_x \le t)$, for varying $t$. Now, in terms of events,
$$ \{ T_x \le t \} \Leftrightarrow \{ T \le t + x \},$$
by definition of $T_x$, and so we know that $P(T_x \le t) = P(T \le t + x)$. But you know the distribution of $T$.
I'll stop here and see if you can finish (it's very nearly done). I can always add more details, should you require. :) -- I will give you a hint to remember that $T$ only takes values in $(0,b)$, so you need to be a little careful if $t > b-x \Leftrightarrow t+x > b$.

You want to look at $P(T_x \le t \mid T > x)$. You can use Bayes' formula/theorem to get this as
$$P(T_x \le t \mid T > x) = \frac{P(T_x \le t, T > x)}{P(T > x)} = \frac{P(x < T < t+x)}{P(T>x)} = \frac{P(T<t+x) - P(T<x)}{P(T>x)}.$$
Now substitute these for the numbers that you know, since you know the distribution of $T$.
In essence this question hinges on the correct application of Bayes' formula, and that $\{ T_x \le t \} \Leftrightarrow \{ T \le t + x \}$.
