Mean of the square of a sum of exponential terms

In summary: If all ##G_k## are independent and identically distributed, then that expectation is zero, and the second term disappears.
  • #1
grepecs
17
0

Homework Statement


[/B]
Calculate [itex]\widehat{Y^{2}}[/itex]

(i.e., the mean of the square of [itex]Y[/itex].

Homework Equations



[tex]Y=\sum_{k=0}^{N-1}y_{k}[/tex]

where

[tex]y_{k}=e^{-\gamma t}e^{\gamma \tau k}G_{k}[/tex]

and

[tex]t=N\tau[/tex]

The quantities [itex]y_{k}[/itex] (or [itex]G_{k}[/itex]) are statistically independent.

The Attempt at a Solution



[tex]\widehat{Y^{2}}=\widehat{G^{2}}e^{-2\gamma t}\sum_{k=0}^{N-1}e^{2\gamma \tau k}=\widehat{G^{2}}e^{-2\gamma t}(\ \frac{1-e^{2\gamma t}}{1-e^{2\gamma \tau}} ) [/tex]

However, the correct answer should be

[tex]\widehat{Y^{2}}=\widehat{G^{2}}\frac{1}{2\gamma\tau} (\ 1-e^{-2\gamma \tau} )[/tex]

so it seems like I'm doing something wrong, or is it possible to somehow simplify my answer in order to get the correct one?
 
Physics news on Phys.org
  • #2
grepecs said:

Homework Statement


[/B]
Calculate [itex]\widehat{Y^{2}}[/itex]

(i.e., the mean of the square of [itex]Y[/itex].

Homework Equations



[tex]Y=\sum_{k=0}^{N-1}y_{k}[/tex]

where

[tex]y_{k}=e^{-\gamma t}e^{\gamma \tau k}G_{k}[/tex]

and

[tex]t=N\tau[/tex]

The quantities [itex]y_{k}[/itex] (or [itex]G_{k}[/itex]) are statistically independent.

The Attempt at a Solution



[tex]\widehat{Y^{2}}=\widehat{G^{2}}e^{-2\gamma t}\sum_{k=0}^{N-1}e^{2\gamma \tau k}=\widehat{G^{2}}e^{-2\gamma t}(\ \frac{1-e^{2\gamma t}}{1-e^{2\gamma \tau}} ) [/tex]

However, the correct answer should be

[tex]\widehat{Y^{2}}=\widehat{G^{2}}\frac{1}{2\gamma\tau} (\ 1-e^{-2\gamma \tau} )[/tex]

so it seems like I'm doing something wrong, or is it possible to somehow simplify my answer in order to get the correct one?

Unless the quantities ##G_k## have mean zero (##E G_k = 0##) and are identically distributed for different ##k## (or, at least, all have the same second moments), the expression you give is incorrect; and in that case the one given in the book (or wherever) is also incorrect. As far as I can see, the only way you could obtain something like the second expression would be to have non-zero values of ##E G_j G_k## of the form ##E G_j G_k = E(G^2) c_{jk}##, where ##E(G^2) \equiv E(G_1^2) = E(G_2^2) = \cdots = E(G_{N-1}^2)##. This is because we have ##Y = \sum_{k=1}^{N-1} w_k G_k## (with ##w_k = e^{-\gamma t} e^{\gamma \tau k}##), so that
[tex] Y^2 = \sum_{k=1}^{N-1} w_k^2 G_k^2 + 2 \sum_{1 \leq j < k \leq N-1} w_j w_k G_j G_k, [/tex]
so you would need an appropriate value for the expectation of the second summation above, and that would depend on the form of the coefficients ##c_{jk}##.
 
  • Like
Likes grepecs

Related to Mean of the square of a sum of exponential terms

1. What is the meaning of the square of a sum of exponential terms?

The square of a sum of exponential terms refers to a mathematical expression in which two or more exponential terms are added together and then squared.

2. How is the mean of the square of a sum of exponential terms calculated?

The mean of the square of a sum of exponential terms is calculated by taking the sum of the squared values of each exponential term, dividing it by the total number of terms, and then taking the square root of the result.

3. What is the significance of calculating the mean of the square of a sum of exponential terms?

Calculating the mean of the square of a sum of exponential terms is useful in many fields, including statistics, physics, and economics. It can help determine the average value of a set of data and can be used to make predictions and analyze trends.

4. Can the square of a sum of exponential terms be negative?

No, the square of a sum of exponential terms cannot be negative. When squared, a number will always result in a positive value.

5. Are there any real-life applications of the mean of the square of a sum of exponential terms?

Yes, the mean of the square of a sum of exponential terms is commonly used in financial analysis, such as calculating risk and return on investment. It is also used in physics to determine the average energy of a system.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
318
  • Calculus and Beyond Homework Help
Replies
1
Views
750
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
549
  • Calculus and Beyond Homework Help
Replies
3
Views
497
  • Calculus and Beyond Homework Help
Replies
1
Views
302
  • Calculus and Beyond Homework Help
Replies
1
Views
496
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
404
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Back
Top