1. ## 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?

2. ## Re: Probability Distribution

"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.

3. ## Re: Probability Distribution

I will show you my working.
I have it on paper will type it out.

4. ## Re: Probability Distribution

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.

5. ## Re: Probability Distribution

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.

6. ## Re: Probability Distribution

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?

7. ## Re: Probability Distribution

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

8. ## Re: Probability Distribution

got it now..
but it will be -2 then..as 0-2=-2
like subtracting from when x approaches infinite minus x approaches zero.

9. ## Re: Probability Distribution

After by part,
when you integrating ,
you should get 1 more negative sign.

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