# Conditional distribution of uniform random variable distributed over (0,1)

Let $U$ be a random variable uniformly distributed over (0,1). Compute the conditional distribution of U given that $U>a$

The solution says:

$P(U > s | U > t) = \frac{P(U > s)}{P(U > t)}$

Here I'm confused why the conditional probability translates into the fraction on the right hand side.

$P(U > s | U > t) = \frac{P(U > s)}{P(U > t)}$

$=\frac{1-s}{1-t}$ where $t \lt s \lt 1$

The distribution function of a uniform variable $P(U \leq x) = \frac{x-a}{b-a}$, so in this case, $P(U > s)=\frac{s-t}{1-t}$ because $t \lt s \lt 1$. $P(U > t)=\frac{t-0}{1}=t$ so $\frac{P(U > s)}{P(U > t)}=\frac{\frac{s-t}{1-t}}{t}$ however this is different from the answer in the solutions, which is $\frac{1-s}{1-t}$. Can someone please explain why my answer is incorrect?

• You seem to be using "$a$" simultaneously to represent two different quantities: both the variable $a$ in the question and the endpoint $0$. That's got to be confusing ... . – whuber Jun 10 '14 at 19:46
• You made exactly the same error again with your edits! In particular, $P(U\gt s) = 1-s$ when $0\le s\le 1$. There really isn't a statistical question here, because it will go away when you get your notation straightened out. – whuber Jun 10 '14 at 20:03

You need to pull $P(U>s)$ from the original distribution, not from the limited range of $(t,1)$. Therefore, $P(U>s) = \frac {1-s}{1-0}$. It's the length of the segment $[s,1]$ divided by the length of the whole segment. We then divide by the probability of the condition $P(U>a) = \frac {1-a}{1-0}$ and you'll have your final answer.

The key thing in conditional probability is that we pull the probabilities from the original distribution, not the new distribution based on the condition. We're actually calculating the new distribution based on the condition.

A concrete example using the same range:

What's the probability that U is larger than $\frac 34$ given that U is larger than $\frac 12$?

We can tell (relatively intuitively) that this will be equal to $\frac 12$. $\frac 34$ is halfway between $\frac 12$ and $1$. But to calculate directly using the formula:

$$P(U>\frac34|U>\frac12) = \frac {P(U>\frac34)}{P(U>\frac12)}$$

From the original range, $P(U>\frac34) = \frac14$ and $P(U>\frac12) = \frac12$. So:

$$P(U>\frac34|U>\frac12) = \frac{\frac14}{\frac12} = \frac12$$

Again, just make sure you're pulling the top probability from the original distribution.

By Bayes,

$$P(U > s | U > t) = \frac{P\left[\{U > s\}\cap \{U > t\} \right]}{P(U > t)}$$

But since by the premises, $s>t$, it follows that $P\left[\{U > s\}\cap \{U > t\} \right]=P(U > s)$. Hence,

$$P(U > s | U > t) = \frac{P(U > s)}{P(U > t)} = \frac {1-s}{1-t}$$

the ratio of two marginal (unconditional) probabilities from $U(0,1)$.

• I'm not sure if it's by Bayes or if it's by the definition of conditional probability, but this is how I would do it. – user44764 Jun 10 '14 at 20:10
• @Matthew thanks. The mention of Bayes was in order to refer to the widely familiar expression that links conditional, unconditional, and joint probability. – Alecos Papadopoulos Jun 10 '14 at 20:15