Cant understand this integral(probability)

  • Thread starter Thread starter nhrock3
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around understanding an integral related to probability, specifically focusing on the integration variable and the interpretation of the limits involved. Participants are examining the relationship between the variables u and v within the context of probability density functions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the implications of integrating with respect to u and questioning the limits of v, particularly why v is constrained between 0 and 1. There is also discussion about the geometric interpretation of areas under curves and their relation to probability.

Discussion Status

The discussion is active, with participants providing different interpretations of the integral and the areas involved. Some guidance on the relationship between the variables has been offered, but there is no consensus on the interpretation of the integral or the areas calculated.

Contextual Notes

There are mentions of a photo that contains unclear markings, which may be affecting participants' understanding. Additionally, there are references to specific areas and their corresponding probabilities, indicating that the problem may involve visual elements that are not fully described in the text.

nhrock3
Messages
403
Reaction score
0
2ngr2uo.jpg


if we integrate by U
then if we take steady point on U axes we
encounter 1-v and 3-v lines
thus making our intervals for each steady U point
but why 0<v<1??
there is a red point in the photo in our area in which v=3

??
 
Physics news on Phys.org
the blue line should be parallel to the horizontal axis
 
since you are integrating in terms of u. for v between 0 and 1, du will be between the two lines mentioned. for v between 1 and 3, u will vary between the v-axis and 3-v. to check it try performing the volume integral which shows that it is indeed a probability density.
 
no if we go by du then we go by the u axes
 
I don't understand the chicken scratch on the photo you posted. The first integral doesn't make sense. The area between 0 and v=1 and u=1 is 1/2*base*height = 1/2*1*1 = 1/2. The probability in this area is 0.

The second shaded area has area of 1/2*3*3-1/2 = 4. The total probability in this area is 1/4*4 = 1 which is your state space.

If you are integrating along u that means the area of a little strip is height*du, the height being delta v
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
5
Views
2K
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 16 ·
Replies
16
Views
5K