Statistics Problem - Uniform Distribution

1. Dec 9, 2013

planauts

Hi,

The question is: http://puu.sh/5GX2G.jpg [Broken]

http://puu.sh/5GX2G.jpg [Broken]

I am not exactly sure what the question is asking.

Here is the answer/solution: http://puu.sh/5GX68.png [Broken]
But I am not sure what is going on.

Could someone please explain what exactly the question is asking, I can figure out the rest.

Thanks,

Last edited by a moderator: May 6, 2017
2. Dec 9, 2013

MrAnchovy

It's asking you to work out the cumulative distribution function (CDF) F(x). This is defined as the probability that a variate X takes on a value less than or equal to a number x, in other words F(x) = P(X ≤ x).

Now in this problem X = max(X1, X2, X3, X4), so for any x in order for X ≤ x to be true we must have X1≤ x and X2≤ x and X3≤ x and X4≤ x.

Are you OK from there?

3. Dec 9, 2013

planauts

Assuming that, I can get the rest of the problem. Once you get, F(x), you can take the derivative to get f(x). To get expected you integrate x*f(x) from 0,1 and x^2 * f(x) for the variance.

However, I am still confused for the first part (i.e. the cumulative function). The max seems to throw me off. There is another problem similar to this one, except it uses min.

http://puu.sh/5HsfD.jpg [Broken]

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The max/min seems to throw me off, what exactly does that represent? :S

Thanks

Last edited by a moderator: May 6, 2017
4. Dec 9, 2013

mathman

X = max(X1, X2, X3, X4) ≤ x is equivalent to all Xi ≤ x.
Since the Xi are independent, P(X≤x) = P(X1≤x)P(X2≤x)P(X3≤x)P(X4≤x) = x4.

5. Dec 9, 2013

planauts

How would you do min?

6. Dec 9, 2013

Office_Shredder

Staff Emeritus
You need to calculate
$$P(X_1 \geq x, X_2 \geq x, X_3 \geq x \text{ and } X_4 \geq x )$$
Can you figure out what this is?