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View Full Version : Averbach & Mehta: Chapter 2, Random Variables: Ex. 2.2 -

ashlee_j501
04-27-2003, 03:50 PM
I need LOTS of help!!!! In case you don't have the book, here is the problem:

"Let X be a r.v. with mixed density, for any alpha, 0 &lt; alpha &lt; 1.

fm(x) = alpha(1/2) if x = 1,2
(1- alpha)x/8 if 3&lt;=x&lt;=5
alpha = 1/2. Here fd(x) = 1/4, for x = 1,2 and fc(x) = x/16 for 3&lt;=x&lt;=5."

Unfortunately I don't know how to include the graphs...sorry.

They then say that Fm(x) = (x^2 + 7) / 32.

My question is how do they get Fm(x)????????? I don't know where the 7 came from.

Ashlee

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Gandalf
04-27-2003, 05:13 PM
They better just be saying that Fm(x) = (x^2+7)/32 for 3&lt;=x&lt;=5, though it also happens to be true for x=1 (but not x=2).

Fm(x) is the probability that the random variable is less than x.

The probability that x = 1 is 0.25 and that x = 2 is 0.25.

For 3&lt;=x&lt;=5 the probability that the random variable is in the interval [3,x] is the integral of fc(t) from 3 to x (where t is just the arbitrary variable of integration. You could integrate f(r)dr, f(s)ds, whatever; I chose f(t)dt). That integral is just (x^2 - 9) / 32. So the total probability that the random variable is less than x is the probability that it is equal to 1 or 2, plus the probability that it is in the interval [3,x].

1/4 + 1/4 + (x^2 - 9)/32 = 16/32 + (x^2-9)/32 = (x^2 + 7)/32.

It is purely coincidence that Fm(1) = Probability (random variable &lt;=1) happens to equal (1^2 + 7)/32.

ashlee_j501
04-27-2003, 05:56 PM
much clearer now!!!!!!!!!!!!!!!!!