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Science Forum Index » Mathematics Forum » Probability with pdf..
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| mina_world |
 Posted: Tue Apr 29, 2008 3:33 am |
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Hello teacher~
Choose three numbers x_1, x_2, x_3 on interval [0, 1] independently.
Let X be the maximum value of {x_1, x_2, x_3}.
Find the probability density function f(x) of X.
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P(X <= x) = P(x_1 <= x and x_2 <= x and x_3 <= x)
= P(x_1 <= x)*P(x_2 <= x)*P(x_3 <= x) by independence.
P(x_1 <= x) = int{0 to x} 1 dx = x
P(x_2 <= x) = int{0 to x} 1 dx = x
P(x_2 <= x) = int{0 to x} 1 dx = x
so, P(X <= x) = x^3.
so, f(x) = (d/dx)x^3 = 3x^2. |
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| hagman |
| Posted: Tue Apr 29, 2008 3:33 am |
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On 29 Apr., 10:33, "mina_world" <mina_wo...@hanmail.net> wrote:
Quote: Hello teacher~
Choose three numbers x_1, x_2, x_3 on interval [0, 1] independently.
Let X be the maximum value of {x_1, x_2, x_3}.
Find the probability density function f(x) of X.
----------------------------------------------------
P(X <= x) = P(x_1 <= x and x_2 <= x and x_3 <= x)
= P(x_1 <= x)*P(x_2 <= x)*P(x_3 <= x) by independence.
P(x_1 <= x) = int{0 to x} 1 dx = x
P(x_2 <= x) = int{0 to x} 1 dx = x
P(x_2 <= x) = int{0 to x} 1 dx = x
so, P(X <= x) = x^3.
In other words:
The probability that the max of three is less than x
is the probability that all three are less than x,
which is x^3.
Quote:
so, f(x) = (d/dx)x^3 = 3x^2.
This clearly generalizes to "max of n" with
density function f(x) = n*x^(n-1). |
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| Ray Vickson |
| Posted: Tue Apr 29, 2008 6:50 am |
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On Apr 29, 1:33 am, "mina_world" <mina_wo...@hanmail.net> wrote:
Quote: Hello teacher~
Choose three numbers x_1, x_2, x_3 on interval [0, 1] independently.
You also need to specify the distribution of the individual x_i (not
just that they are independent). Do you mean they are chosen /
uniformly/? If that is what you mean, you need to learn to say it.
Quote:
Let X be the maximum value of {x_1, x_2, x_3}.
Find the probability density function f(x) of X.
----------------------------------------------------
P(X <= x) = P(x_1 <= x and x_2 <= x and x_3 <= x)
= P(x_1 <= x)*P(x_2 <= x)*P(x_3 <= x) by independence.
P(x_1 <= x) = int{0 to x} 1 dx = x
P(x_2 <= x) = int{0 to x} 1 dx = x
P(x_2 <= x) = int{0 to x} 1 dx = x
so, P(X <= x) = x^3.
It is not clear what you want, since you did not ask a question. Did
you want us to confirm that you are correct?
All this is standard, Probability 101 material. If the X_i have common
density f(x), with F(x) = int[f(y),y=0..x] being the cumulative
distribution of each X_i, then Z = max(X_i,i=1,...,n) has cdf F(x)^n,
so has density f_Z(x) = n * F(x)^(n-1) * f(x). The minimum Y =
min(X_i,i=1,...,n) has complementary cumulative distribution P{Y > x}
= [P{X > x}]^n = [1-F(x)]^n, so Y has density f_Y(x) = -(d/dx) [1 -
F(x)]^n = n * f(x) * [1-F(x)]^(n-1). For the uniform distribution case
we have f(x) 1 and F(x) = x on [0,1].
R.G. Vickson
Quote:
so, f(x) = (d/dx)x^3 = 3x^2. |
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