Originally posted by Ahsan89
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Max and Min order statistics probabilities
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Originally posted by Ahsan89 View PostHello everyone .... i've got a very simple problem but my answer doesn't seem to match to that problems solution ...... just wanted to check whether i am missing something or that solutions not correct
f(x) = {x^2 1}/4 2<x<2
E[X] = ?
when i solved it i got 1 but solutions saying that its 0
any help would be appreciated.thanks
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Thanks for your response ... Heres what i did
E[X] = {int(2,0)} (x)(x^21)/4 dx + {int(0,2)} (x)(x^21)/4 dx
= 1/4 [ {int(2,0)} (x^3  x) dx }] + 1/4[{int(0,2)} (x^3  x) dx }]
=1/4 [(x^4)/4 + (x^2)/2}limits(2,0)] + 1/4 [ {(x^4)/4  (x^2)/2}limits(0,2)
=1/4 [0+0 +(2)^4/4  (2)^2/2] + 1/4[ (2)^4/4 (2)^2/2 0+0]
=1/4 [4  2] + 1/4 [4  2]
= 1/2 + 1/2
= 1
Ain't this correct or did i miss something ?
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Apollon is correct, if you recognize this as a single random variable with a pdf that's symmetric around the Y axis, you can save yourself the trouble of calculation.
If you want the comfort of performing the math, you'll need 3 integrals:
Does the image help?
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Part of the problem was getting it set up correctly. You have it set up above as:
E[X] = {int(2,0)} (x)(x^21)/4 dx + {int(0,2)} (x)(x^21)/4 dx
Which looks like you expected everything to the left of the Y axis to be "flipped" by the absolute value. If that were correct then you still would have needed to put a minus sign in front of the portion that was flipped, i.e.
I'm not suggesting that this is the solution to the problem you've stated, but supposing it were you would have wanted something like this:
E[X] = {int(2,0)} (x)(x^21)/4 dx + {int(0,2)} (x)(x^21)/4 dx
Note that this will return a zero, but it's only incidentally be the correct answer.
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