Expected bounds of a continuous bi-variate distribution

[transmini]

Homework Statement

[/B]
$-1\leq\alpha\leq 1$

$f(y_1,y_2)=[1-\alpha\{(1-2e^{-y_1})(1-2e^{-y_2})\}]e^{-y_1-y_2}, 0\leq y_1, 0\leq y_2$
and $0$ otherwise.

Find $V(Y_1-Y_2)$. Within what limits would you expect $Y_1-Y_2$ to fall?

Homework Equations

N/A

The Attempt at a Solution

[/B]
I understand how to go about getting the variance of this distribution. That's not a problem. What I don't understand is finding the expected limits of $Y_1-Y_2$. The book has the solution as $\mu_{Y_1-Y_2} \pm 2*\sigma_{Y_1-Y_2}$. I can't find anything about this in my book with what's been covered thus far in this course or the last course. 2 standard deviations just seems rather arbitrary in this case. Is there a reasoning for 2 standard deviations? Possibly because the marginal distribution functions are exponential distributions, or because there's some convention to use 2 instead of say 1 or 3?

Note: This may actually be covered in the future weeks, as this book likes to use material from future sections in questions of previous sections. For example, in this question, the solutions in the back use the covariance to find the variance of $Y_1-Y_2$, whereas covariance isn't introduced until the next section.

 

[Ray Vickson]

Homework Statement

[/B]
$-1\leq\alpha\leq 1$

$f(y_1,y_2)=[1-\alpha\{(1-2e^{-y_1})(1-2e^{-y_2})\}]e^{-y_1-y_2}, 0\leq y_1, 0\leq y_2$
and $0$ otherwise.

Find $V(Y_1-Y_2)$. Within what limits would you expect $Y_1-Y_2$ to fall?

Homework Equations

N/A

The Attempt at a Solution

[/B]
I understand how to go about getting the variance of this distribution. That's not a problem. What I don't understand is finding the expected limits of $Y_1-Y_2$. The book has the solution as $\mu_{Y_1-Y_2} \pm 2*\sigma_{Y_1-Y_2}$. I can't find anything about this in my book with what's been covered thus far in this course or the last course. 2 standard deviations just seems rather arbitrary in this case. Is there a reasoning for 2 standard deviations? Possibly because the marginal distribution functions are exponential distributions, or because there's some convention to use 2 instead of say 1 or 3?

Note: This may actually be covered in the future weeks, as this book likes to use material from future sections in questions of previous sections. For example, in this question, the solutions in the back use the covariance to find the variance of $Y_1-Y_2$, whereas covariance isn't introduced until the next section.

Is this question copied exactly as it was given to you? The problem is that the notion of "expected limits" is not a standard statistical concept. It may be that the word "expected" is being used in this case in its ordinary, everyday meaning, rather than in its technical statistical/probabilistic sense. If so, the whole thing is too vague to be of much use. If it wants you to find an interval $(a,b)$ such that $P(a < Y_1 - Y_2 < b) \geq p_0$ for some specified numerical value $p_0$, it should just say that.

 

[transmini]

Is this question copied exactly as it was given to you? The problem is that the notion of "expected limits" is not a standard statistical concept. It may be that the word "expected" is being used in this case in its ordinary, everyday meaning, rather than in its technical statistical/probabilistic sense. If so, the whole thing is too vague to be of much use. If it wants you to find an interval $(a,b)$ such that $P(a < Y_1 - Y_2 < b) \geq p_0$ for some specified numerical value $p_0$, it should just say that.

Yeah, this is exactly as it was given, word for word. There are 2 or 3 parts before this, but they were all problems of the "find this expected value" or "find this variance" type. So they shouldn't be relevant. In my own opinion, I don't think this text is all that great because they frequently require future material or give vague questions like this.