Tricky math stats

[Mathstudent7]

a and b are two different components in an electronic system. Component a has durability of X and component b has a durability of Y. X and Y are independent random variables. Furthermore X is uniform with [0, 10] in months and Y is exponential distributed with expected durability 11 months. What's the probability that a breaks before b. Ie P[X<Y]

I know that because they are independent:
f(x, y) = f(x)f(y)
Ie I can get the probabilty function for f(x, y) but I have no idea what to do with it.

 

[Deleted member 4993]

a and b are two different components in an electronic system. Component a has durability of X and component b has a durability of Y. X and Y are independent random variables. Furthermore X is uniform with [0, 10] in months and Y is exponential distributed with expected durability 11 months. What's the probability that a breaks before b. Ie P[X<Y]

I know that because they are independent:
f(x, y) = f(x)f(y)
Ie I can get the probabilty function for f(x, y) but I have no idea what to do with it.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[Mathstudent7]

I know that because they are independent:

f(x, y) = f(x)f(y) = 1/110 exp (-y/11)

From here I get confused. I'm considering taking integral of f(x, y) dx dy. But I don't know which limits to use, if this even is the right approach.

Because the joint probability distribution F(x, y) = P [ Y<y, X < x] which is not what i want, rather I am looking for P[X<Y]

 

[Dr.Peterson]

I know that because they are independent:

f(x, y) = f(x)f(y) = 1/110 exp (-y/11)

From here I get confused. I'm considering taking integral of f(x, y) dx dy. But I don't know which limits to use, if this even is the right approach.

Because the joint probability distribution F(x, y) = P [ Y<y, X < x] which is not what i want, rather I am looking for P[X<Y]

Try integrating over the region where x<y. (Sketch a graph of that region first, and pay attention to the domain.)

 

[Mathstudent7]

Intuitively I am thinking x<y when 1/10 < 1/11 exp(-y/11) => y < -1.0484 but that doesnt make sense because 0<y<infinity for the exponential distribution of Y

 

[Dr.Peterson]

This is not a matter of solving for y, but of setting the limits of the integral.

What would you do to find the area of the region defined by [MATH]x < y[/MATH], [MATH]0 \le x \le 10[/MATH], and [MATH]y \ge 0[/MATH]? Do the same sort of thing here.

 

[Mathstudent7]

Thanks. I used the limits zero to infinity for y, and zero to y for x. The problem is when I integrate this the result is infinity which should not be possible. please see the attached file

 

[Dr.Peterson]

You forgot that x must be less than 10.

When I sketch the region, it is obvious that I want to integrate with respect to y first, not x. If you do as you did, then you need to break it into two regions, above and below y=10.

Show me your sketch of the region; maybe you made a mistake there.

 

[Mathstudent7]

Think I (you) solved it now, please see if the solution is correct. Thanks a lot for your help Dr. Peterson. For others new to double integrals a recommend to read this:

In double integral, does it matter which variable is integrated first?

Answer (1 of 5): To compute a double integral, one cannot in general change the order of integration. As explained in many answers, changing the order of integration obviously changes the bounds. But the core of the problem is that the iterated integration method (ie, integrate x first, then y or...

www.quora.com

EDIT: a bit confused again: does this implie that Y most be : y >10
despite the premiss of the question that y>0. Because if x is 0->10 and y>x does this mean that y must be greater than 10?

 

[Dr.Peterson]

Your work is correct now.

I'm not sure what your question about y>10 means. The region over which you are integrating includes values of y > 10, but not only those values. The fact that x can reach 10 and y is greater doesn't mean that y must be greater than 10.

If you integrated with respect to x first, then you'd need one integral over y from 0 to 10, and another for y > 10.

Since you never showed the region over which you are integrating, I will:



You are working with the purple region. Clearly y is not always greater than 10 in that region; but its behavior changes at y = 10.