i need help with finding chebyshev inequality, please

Reply Thu 3 Nov, 2011 10:01 am
a random variable X has pdf

f(x)=(3/2)e^-|3x| -∞ < x < ∞

a. find the mean and variance of X
b. use the chebyshev inequality to find a number P(|X-μx>α)=.01

i try to integrate (3/2)xe^-|3x|dx from -∞ to ∞ but keep getting zero for the mean.
should i split -|3x| into -3x and 3x then integrate (3/2)e^-(3x) and (3/2)e^3x .
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Reply Fri 4 Nov, 2011 04:16 am
The average value of a function is the integral divided by the range it's integrated over.

From your description I have a couple of problems. If the function is

f(x)=(3/2)e^-|3x| over the range from -∞ < x < ∞

Why are you integrating

f(x)=(3/2)xe^-|3x|dx where did the x come from?

second look at the absolute value

isn't |-3x|=|3x|

So isn't the integral of f(x)=(3/2)e^-|3x| over the range from -∞ < x < ∞
equal to twice the integral of f(x)=(3/2)e^-3x over the range from 0 < x < ∞

Hope that's some help.


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Reply Wed 22 Aug, 2012 06:14 am
You made no mistakes , friend . f(x) is the p.d.f. (I in fact verified it is a p.d.f.) so mean is the integration of xf(x) over domain of definition of x
which in this case is from - infinity t0 + infinity ; now xf(x)=(3/2)xe^ -l3xl is an odd function so the integration must yield o , and it is the mean . It is not unusual to have o as the mean for a p.d.f. , as you might know the standard-normal distribution , the standard Cauchy distribution etc. have mean 0
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