Question

HEY EXPERTS! If I am calculating the expected value of a continuout random variable X, and I have F(x) = {three piece wise function here}, do I use two integrals? That is, the function are 0 when X <= -1, x^2 when 0 <= X <= 4 and 1 when x > 4. I know one i setup a definite integral w/ limits from 0 to 4, but how would i setup the integral from 4 to infinity?

Answer 1

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Answer 2

@tkhunny @Ashleyisakitty @robtobey @zepdrix @Mr.Man420 @praxer @iPwnBunnies @NY,NY @courter98 @Ilovecake @vshiroky @halorazer

Answer 3

\[E(X) = \int_{\-infinity}^{\infinity} xf(x)dx \]

Answer 4

man you are all worthless

Answer 5

Are you new to OpenStudy?

Answer 6

@prodemis

Answer 7

yes y

Answer 8

Tagging more than about 4 people is not allowed. I recommend that you read the CoC.
http://openstudy.com/code-of-conduct
If no one answers your question, then you may bump it after a period of time, and when someone has the knowledge to help, they will. It is also not very nice to call people on here worthless. Most of the people on here are very nice and help many people. Please be nicer and follow the rules in the future.

Answer 9

You are confusing F(x) and f(x). ALL of the probability in this distribution is in [0,4]. Therefore, \(\int\limits_{-\infty}^{\infty}xf(x)\;dx = \int\limits_{0}^{4}xf(x)\;dx\)

HOWEVER, there are a few holes in this definition.
1) [-1,0] seems to be missing.
2) With F(x) defined as \(x^{2}\), this works for \(x\in[0,1]\), or some other much smaller Domain. It is certainly MUCH too large on [0,4]

Please provide a reasonable definition. THEN, someone might be able to answer your questions.

Answer 10

Your worthless

Answer 11

You could learn how to spell. That probably would help.