# Homework Help: Continuous uniform distribution function

1. Feb 3, 2017

### theone

1. The problem statement, all variables and given/known data
Can someone explain why f(x) = 1/(b-a) for a<x<b ?

2. Relevant equations

3. The attempt at a solution
shouldn't it be 0? since its a continuous random variable and so that interval from a to b has an infinite number of possible values?

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2. Feb 3, 2017

### Diegor

The probability to get a particular value for X is 0 but if you want to know the probability to get X within a range, for example 3<X<4 you need to integrate the probability density between these two values and that is not always 0. In a continuous distribution f(x) is a probability density. In the case of the example you posted it means you should integrate to get probability. Try to integrate f(x) and you will get interesting results, you may also understand why f(x) has that form.

3. Feb 3, 2017

### Ray Vickson

What do YOU think $f(x)$ means in this context? (Perhaps you are mis-interpreting the symbols, etc.)

4. Feb 3, 2017

### theone

the probability that a continuous random variable X takes on one of its possible values x?

5. Feb 3, 2017

### theone

to get the probability of x falling within a range, aren't you essentially adding the probabilities of x taking on the particular values within the range... but if the probability of x taking on a particular value is 0, then why is this sum not always zero?

6. Feb 3, 2017

### Ray Vickson

No, absolutely not. For a continuous random variable the probability that $X = x$ is zero for all $x.$ However, the probability that $X$ takes a value between $x$ and $x+\Delta x$ (for small but nonzero $\Delta x$) is given by $f(x) \Delta x$ (to lowest order in $\Delta x$. For a uniform distribution on the interval $[a,b]$, we have
$$P\{ c < X < d \} = \frac{d-c}{b-a} = \frac{\text{small length}}{\text{large length}}$$
for any $a \leq c < d \leq b$.

In other words, for a uniform distribution over an interval the probability of a sub-interval is proportional to the length of that sub-interval, but is independent of its "location". We get the same probability whether the sub-interval is near the beginning, in the middle, or near the end of the main interval.

7. Feb 3, 2017

### Ray Vickson

Because we have an uncountably infinite set of points. The probability of an interval is NOT a SUM of point probabilities!

If I take the interval from 0 to 1 would you say its length must be zero because it is made up of points having length zero? If I take a triangle of base 2 and height 1 would you say its area is zero because it is made up of points of area zero? Well, probability behaves just like length or area in that regard.

8. Feb 3, 2017

### Diegor

That's why integration is involved and not regular sum like in discrete distributions. (Trying to adapt to discrete would be something like p =succes cases/posibble cases = infinity/infinity)

Think this way: Suppose that the probability to find values of x in a range let's say between 3 and 4 is not zero and then you take a small interval like 3,5 to 3,51, you will see that the probability is smaller. If you continue to shrink the interval you will find that the probability tends to zero but still you have a non zero probability in the (3,4) interval.

For example you go to the market to buy one liter of milk. It is unlikely that you find one bottle with exactly one liter but within 0,9 to 1,1 liters surely you will get more than one.