Exam P Question for my Probability Homework

[pinky14]

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.$$$$

 

[tkhunny]

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. :-)

 

[pinky14]

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.

 

[HallsofIvy]

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".

 

[pinky14]

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.

 

[tkhunny]

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!