suppose we have following regression model
basic question is : if we change (increase or decrease ) any variable by 5 percentage , how it will affect on y variable?i think first we should change given variable(increase or decrease by 5 percentage ) first and then sketch regression , estimate coefficients of corresponding variable and this will answer, how effect it will be right?and if question is how much percentage of changing we will have, then what we should do? thanks in advance
you are right-Betas are noting but the amount of change in Y, if a unit of independent variable changes.( Keeping other X constant)
http://www.theanalysisfactor.com/interpreting-regression-coefficients/
Can't you take % change in Y value when you make % change in X values. I think this will help. Example- if Y changes from 20 to 25 , you can say it has increased by 25%.
Related
I have been given an exercise for my course and could use some help with it. We have to turn a pixel (row x column) into its memory address and print it. $gp is pixel 0x0 and every pixel is 32 bits. How would I go about calculating let's say pixel 0,1?
(width = 32px, height = 16px)
I've looked everywhere in our course information and can't seem to find anything to help me out.
Firstly you do y * width + x = index. Then you have to multiply the index by the size of the pixel to get the offset and finally $gp + offset.
I am having issue in getting clear concept of contrastive loss used in siamese network.
Here is pytorch formula
torch.mean((1-label) * torch.pow(euclidean_distance, 2) +
(label) * torch.pow(torch.clamp(margin - euclidean_distance, min=0.0), 2))
where margin=2.
If we convert this to equation format, it can be written as
(1-Y)*D^2 + Y* max(m-d,0)^2
Y=0, if both images are from same class
Y=1, if both images are from different class
What i think, if images are from same class the distance between embedding should decrease. and if images are from different class, the distance should increase.
I am unable to map this concept to contrastive loss.
Let say, if Y is 1 and distance value is larger, the first part become zero (1-Y), and second also become zero, because it should choose whether m-d or 0 is bigger.
So the loss is zero which does not make sense.
Can you please help me to understand this
If the distance of a negative sample is greater than the specified margin, it should be already separable from a positive sample. Therefore, there is no benefit in pushing it farther away.
For details please check this blog post, where the concept of "Equilibrium" gets explained and why the Contrastive Loss makes reaching this point easier.
I want to limit the max value of the vertical axis to 100 % and not 101%.
I have tried setting the axis limit to 100 but that did not help either. Changing that to 100.05% just changes the axis label but the line does not jump to the top of the chart. Is there a way to restrict the high value of the axis to a particular number?
Follow up question: I see this in ChartDataLabel. What does this mean =IIF(Fields!Reliability.Value >= 100, "", Fields!Reliability.Value) ?
To do what you want, while in design mode, right-click on the axis and choose Vertical Axis Properties. Set the Minimum and Maximum values to what you want.
Now, it sounds like that is something you have already tried, but didn't like the results. Another thing you can do is use the values from the data set, along with an expression, to set the min and max values for the axis. So, instead of setting the min and max values manually, you can try something like the following. This assumes the value you are charting is Fields!Reliability.Value.
Minimum: =Min(Fields!Reliability.Value)
Maximum: =Max(Fields!Reliability.Value)
If you want a little bit of a buffer above and below the value line, then you can add and subtract from those values like this:
Minimum: =Min(Fields!Reliability.Value)*0.95
Maximum: =Max(Fields!Reliability.Value)*1.05
Adjust these as desired.
To answer your follow up question about the ChartDataLabel.
=IIF(Fields!Reliability.Value >= 100, "", Fields!Reliability.Value)
This expression sets the ChartDataLabel to an empty string if the Fields!Reliability.Value from the data set is greater than 100, otherwise it uses whatever value is in Fields!Reliability.Value.
Hope this helps you out.
I've been struggling to calculate the accelerator. I've spend a whole day in searching, trial & error but all in vain. I've one horizontal line on the stage (AS3) of let say 200 width. Center-point of that line is on 60 (if it was 100, I would have surely done it by just calculating the percentage). Now I need to know the width of given percentage. For example, total width of 60% or where will 30% (or any other percentage) start from?
What I know is the total width, and the center-point (either in percentage or in width).
Your help will be highly appreciated. In case if there is any formula, please give me details, don't just mention a/b/c as I'd never been a student of physics :(
Edit:
I don't have 10 reputations, so I can't post image directly here. Please click the following link to see the image.
Link: http://oi62.tinypic.com/11sk183.jpg
Edit:
Here is what I want exactly: I want to travel n% from any point (A/B/C/D) to its relative point (A->B/A->D ...) (Link)
http://i59.tinypic.com/2wp2lbl.jpg
If I understand correctly, you want a non-linear scale, so that pixel 1 on the line is 0%, pixel 100 on the line is 60% and pixel 200 is 100%?
If x=pixelpos/200 is the relative position on the line, one easy variation of the linear scale y=x*100% is y=(x+a*x*(1-x))*100%.
For x=0.5 the value is y=0.5+a*0.25, so for that to be 0.6=60% one needs a=0.4.
To get in the reverse direction the x for y=0.3=30%, one needs to solve a quadratic equation y=x*(1+a*(1-x)) or a*x^2-(1+a)*x+y=0. With the general solution formula, this gives
x = (1+a)/(2*a)-sqrt((1+a)^2-4*a*y)/(2*a)
= (2*y) / ( (1+a) + sqrt((1+a)^2-4*a*y) )
= (2*y) / ( (1+a) + sqrt((1-a)^2+4*a*(1-y)) )
and with a=0.4 and y=0.3
x = 0.6/( 1.4 + sqrt(1.98-0.48) )
approx 0.6/2.6=3/13=231/1001 approx 0.23
corresponding to pixel 46.
This will only work for a between -1 and 1, since for other values the slope at x=0 or x=1 will not be positive.
Another simple formula uses hyperbola instead of parabola,
y=a*x/(1+(a-1)*x)
with the inversion by
y+(a-1)*x*y = a*x <=> y = (a-(a-1)*y)*x
x = (y/a)/(1+(1/a-1)*y)
and
a = (y*(1-x))/(x*(1-y))
here there is no problem with monotonicity as long as there is no pole for x in [0,1], which is guaranteed for a>0.
How can I set the text size (inside TextField) in standart CSS/printable points? According to the manual:
fontSize - Only the numeric part of the value is used. Units (px, pt)
are not parsed; pixels and points are equivalent.
As far as I understand, 1 pixel may be equal to 1 point only in 72 PPI case. So, actionscript just operating pixels (not the real points). My trouble is to get the actual text size that I can print. Any advices or solutions are welcome.
SWF is measured in pixels, moreover, is scalable, so 1 pixel can be 1 point now, 2 points a bit later (scaleY=scaleX=2), and an undefined number another bit later (removed from stage without dereferencing). In short, for AS there are NO "real points" since it does not know a thing about printers, while it knows about displays.