MHB Exam P Question for my Probability Homework

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The discussion focuses on calculating the expected operational time and variance of a machine with two components, using a joint density function. Participants emphasize the need for three integrals: one to verify the probability distribution, one to find the expected value, and another to calculate the variance. The first integral confirms the distribution is valid, while the second and third integrals are used to derive the expected value and variance, respectively. There is a consensus that understanding multivariable calculus is essential for solving these problems effectively. Mastery of these concepts is crucial for success in the course.
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A machine consists of two components whose lifetimes have a joint density function:

$f(x,y) = \left\{
\begin{array}{ll}
\frac{1}{50} & \quad x \geq 0, y \geq 0, x+y \leq 10 \\
0 & \quad Otherwise
\end{array}
\right.$

The machine operates until both components fail.

(a) Calculate the expected operational time of the machine. Hint: This is E(X + Y).

(b) Calculate the variance of the operational time of the machine.$$$$
 
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This will require your best multi-variable calculus.

Can you state the definition of "Expected Value" in this context?

Personally, I would do three (3) integrals to complete this assignment:

With appropriate limits...
1) \int\int f(x,y)\;dy\;dx Just to show that we do have a proper probability distribution.
2) \int\int (x+y)\cdot f(x,y)\;dy\;dx To find the Expected Value
3) \int\int (x+y)^{2}\cdot f(x,y)\;dy\;dx To get us on our way to the Variance.

Let's see where that takes you. :-)
 
Last edited:
tkhunny said:
This will require your best multi-variable calculus.

Can you state the definition of "Expected Value" in this context?

Personally, I would do three (3) integrals to complete this assignment:

With appropriate limits...
1) \int\int f(x,y)\;dy\;dx Just to show that we do have a proper probability distribution.
2) \int\int (x+y)\cdot f(x,y)\;dy\;dx To find the Expected Value
3) \int\int (x+y)^{2}\cdot f(x,y)\;dy\;dx To get us on our way to the Variance.

Let's see where that takes you. :-)

Why are we doing 3 integrals? I thought that when we calculate expected value, it is just E(X) or in this case E(X+Y) and that when we calculate variance, we are just doing $E(X^2)-[E(X)]^2$ or in this case Var(X+Y) = Var(X) + 2Cov(X,Y) + Var(Y)? I am honestly not understanding a lot of concepts in my class but I am trying to.
 
Yes, that was what tkhunny said! He suggested you do the first integral just to determine if this is a valid joint probability distribution:
\int_0^{10}\int_0^{10- x} \frac{1}{50} dydx= \frac{1}{50}\int_0^{10} 10- x dx= \frac{1}{50}\left[10x- \frac{1}{2}x^2\right]_0^10= \frac{1}{50}(100- 50)= \frac{50}{50}= 1.

Or, more simply, with 0\le x\le 10 and 0\le y \le 10 we have a 10 by 10 square with area 100. But the probability is non-zero only for x+ y\le 10, the lower triangular half of the square which has area 50. The probability distribution there is the constant 1/50. 50(1/50)= 1. Since the "total probability" is 1 this is a valid joint probability distribution.

The second integral, \int\int (x+ y)f(x, y)dy dx is the expected value of x+ y. That is \frac{1}{50}\int_0^{10}\int_0^{10- x} x+ y dy dx= \frac{1}{50}\int_0^{10}\left[10y- \frac{1}{2}y^2\right]_0^{10- x}dx. Complete that calculation.

The third integral, \frac{1}{50}\int_0^{10}\int_0^{10- x}(x+ y)^2 dy dx, is the "variance".
 
HallsofIvy said:
Yes, that was what tkhunny said! He suggested you do the first integral just to determine if this is a valid joint probability distribution:
\int_0^{10}\int_0^{10- x} \frac{1}{50} dydx= \frac{1}{50}\int_0^{10} 10- x dx= \frac{1}{50}\left[10x- \frac{1}{2}x^2\right]_0^10= \frac{1}{50}(100- 50)= \frac{50}{50}= 1.

Or, more simply, with 0\le x\le 10 and 0\le y \le 10 we have a 10 by 10 square with area 100. But the probability is non-zero only for x+ y\le 10, the lower triangular half of the square which has area 50. The probability distribution there is the constant 1/50. 50(1/50)= 1. Since the "total probability" is 1 this is a valid joint probability distribution.

The second integral, \int\int (x+ y)f(x, y)dy dx is the expected value of x+ y. That is \frac{1}{50}\int_0^{10}\int_0^{10- x} x+ y dy dx= \frac{1}{50}\int_0^{10}\left[10y- \frac{1}{2}y^2\right]_0^{10- x}dx. Complete that calculation.

The third integral, \frac{1}{50}\int_0^{10}\int_0^{10- x}(x+ y)^2 dy dx, is the "variance".

How did you get the limits? I have trouble tying to draw it and know how to see which way of integration would be easier, my calculus is a bit rusty.
 
It's just a triangle. Have you already had multivariable calculus in your studies? You will need it. If you have not studied it, you will have a very difficult time getting through this course.

Integral #1 s/b 1 <== Yea! A valid probability distribution.

Integral #2 s/b Mean, or First Moment

Integral #3 s/b Second Moment. Variance = (Second Moment) - (First Moment)^2

Time to get up to speed!
 

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