Probability Distribution
The time (in hours) it takes before belts X and Y need replacing in a high use machine follows the joint probability distribution:
integral y/100* e^ -x/2 where x>0, 0<y<10
My E(x) comes to be approaching to infinity as x approaches infinity.
is it possible?
"My E(x) comes to be approaching to infinity as x approaches infinity."
What does it mean?
If E[X] is the expected value of X, what is x? the dummy variable?
Once you have taken the expected value, the dummy variable x is integrated
out. It should be independent from your x.
Moreover, since the joint p.d.f. is a product of the functions of the dummy
variables x and y, you should know that X, Y are independent, and X should
have an exponential distribution.
I will show you my working.
I have it on paper will type it out.
g(x)=1/2*e^(-x/2)
E(x)=integral X*g(x) dx x>0
E(x)=integral X* 1/2*e^(-x/2) dx
E(x)=after integration by parts..
I get
=1/2[-2x/e^-x/2+x^2/2*e^-x/2]..
now putting 0 and inifinte
I get
=Infinite.
Hmm. Must be getting that integral wrong somewhere. I couldn't really make out what your final answer was. But you should be getting
and then you plug in the limits and you're should get 2.
I got to this point but isnt it like infinite plus infinite? if we put infinite and then subtract putting the lower limit that is 0?
If we rewrite it as
And then what happens to xe^{-x} as x goes to infinity? If you don't know then rewrite it as:
if you get an indeterminate form you could try L'Hopital
got it now..
but it will be -2 then..as 0-2=-2
like subtracting from when x approaches infinite minus x approaches zero.
After by part,
when you integrating,
you should get 1 more negative sign.
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