Statistics Problem - Uniform Distribution

[planauts]

Hi,

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

http://puu.sh/5GX2G.jpg

I am not exactly sure what the question is asking.


Here is the answer/solution: http://puu.sh/5GX68.png
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,

 

[pbuk]

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?

 

[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

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

Thanks

 

[mathman]

However, I am still confused for the first part (i.e. the cumulative function). The max seems to throw me off.

MrAnchovy gave you the answer.
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.

 

[planauts]

How would you do min?

 

[Office_Shredder]

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?