Q)A box contains 4 red balls, 3 green balls and 3 blue balls. Two balls are selected at
random without replacement. Let X represent the number of red balls in the sample and
Y the number of green balls in the sample.
a) Arrange the different pairs of values of (X, Y ) as the cells in a table, each cell being
filled with the probability of that pair of values occurring, i.e. provide the joint
probability distribution.
b) What does the random variable Z = 2 - X - Y represent?
c) Calculate Cov(X, Y ).
d) Calculate P(X = 1 | -2 < X - Y < 2).
I couldn't understand how to think to solve the part a) in this question and so on.
To solving this question first of all you have to create a tree with this two events. First data in this question is that we can take is , these are not independent event. so you can create tree like this,
In first part you have to create the joint table of X and Y.
there is 0,1,2 are the only possible values that each variable can get.
The critical situations are that X-1 , Y-0 and X-0 , Y-1 .Because they got two possible chances in same situation that one color ball take first and that same color ball select the second time.
So this is the table that can get according to this tree.
part b represent the blue balls in selected sample
I have 3 questions:
1)
The confusion matrix for sklearn is as follows:
TN | FP
FN | TP
While when I'm looking at online resources, I find it like this:
TP | FP
FN | TN
Which one should I consider?
2)
Since the above confusion matrix for scikit learn is different than the one I find in other rescources, in a multiclass confusion matrix, what's the structure will be? I'm looking at this post here:
Scikit-learn: How to obtain True Positive, True Negative, False Positive and False Negative
In that post, #lucidv01d had posted a graph to understand the categories for multiclass. is that category the same in scikit learn?
3)
How do you calculate the accuracy of a multiclass? for example, I have this confusion matrix:
[[27 6 0 16]
[ 5 18 0 21]
[ 1 3 6 9]
[ 0 0 0 48]]
In that same post I referred to in question 2, he has written this equation:
Overall accuracy
ACC = (TP+TN)/(TP+FP+FN+TN)
but isn't that just for binary? I mean, for what class do I replace TP with?
The reason why sklearn has show their confusion matrix like
TN | FP
FN | TP
like this is because in their code, they have considered 0 to be the negative class and one to be positive class. sklearn always considers the smaller number to be negative and large number to positive. By number, I mean the class value (0 or 1). The order depends on your dataset and class.
The accuracy will be the sum of diagonal elements divided by the sum of all the elements.p The diagonal elements are the number of correct predictions.
As the sklearn guide says: "(Wikipedia and other references may use a different convention for axes)"
What does it mean? When building the confusion matrix, the first step is to decide where to put predictions and real values (true labels). There are two possibilities:
put predictions to the columns, and true labes to rows
put predictions to the rows, and true labes to columns
It is totally subjective to decide which way you want to go. From this picture, explained in here, it is clear that scikit-learn's convention is to put predictions to columns, and true labels to rows.
Thus, according to scikit-learns convention, it means:
the first column contains, negative predictions (TN and FN)
the second column contains, positive predictions (TP and FP)
the first row contains negative labels (TN and FP)
the second row contains positive labels (TP and FN)
the diagonal contains the number of correctly predicted labels.
Based on this information I think you will be able to solve part 1 and part 2 of your questions.
For part 3, you just sum the values in the diagonal and divide by the sum of all elements, which will be
(27 + 18 + 6 + 48) / (27 + 18 + 6 + 48 + 6 + 16 + 5 + 21 + 1 + 3 + 9)
or you can just use score() function.
The scikit-learn convention is to place predictions in columns and real values in rows
The scikit-learn convention is to put 0 by default for a negative class (top) and 1 for a positive class (bottom). the order can be changed using labels = [1,0].
You can calculate the overall accuracy in this way
M = np.array([[27, 6, 0, 16], [5, 18,0,21],[1,3,6,9],[0,0,0,48]])
M
sum of diagonal
w = M.diagonal()
w.sum()
99
sum of matrices
M.sum()
160
ACC = w.sum()/M.sum()
ACC
0.61875
I am trying to calculate percentage growth in excel with a positive and negative number.
This Year's value: 2434
Last Year's value: -2
formula I'm using is:
(This_Year - Last_Year) / Last_Year
=(2434 - -2) / -2
The problem is I get a negative result. Can an approximate growth number be calculated and if so how?
You could try shifting the number space upward so they both become positive.
To calculate a gain between any two positive or negative numbers, you're going to have to keep one foot in the magnitude-growth world and the other foot in the volume-growth world. You can lean to one side or the other depending on how you want the result gains to appear, and there are consequences to each choice.
Strategy
Create a shift equation that generates a positive number relative to the old and new numbers.
Add the custom shift to the old and new numbers to get new_shifted and old_shifted.
Take the (new_shifted - old_shifted) / old_shifted) calculation to get the gain.
For example:
old -> new
-50 -> 30 //Calculate a shift like (2*(50 + 30)) = 160
shifted_old -> shifted_new
110 -> 190
= (new-old)/old
= (190-110)/110 = 72.73%
How to choose a shift function
If your shift function shifts the numbers too far upward, like for example adding 10000 to each number, you always get a tiny growth/decline. But if the shift is just big enough to get both numbers into positive territory, you'll get wild swings in the growth/decline on edge cases. You'll need to dial in the shift function so it makes sense for your particular application. There is no totally correct solution to this problem, you must take the bitter with the sweet.
Add this to your excel to see how the numbers and gains move about:
shift function
old new abs_old abs_new 2*abs(old)+abs(new) shiftedold shiftednew gain
-50 30 50 30 160 110 190 72.73%
-50 40 50 40 180 130 220 69.23%
10 20 10 20 60 70 80 14.29%
10 30 10 30 80 90 110 22.22%
1 10 1 10 22 23 32 39.13%
1 20 1 20 42 43 62 44.19%
-10 10 10 10 40 30 50 66.67%
-10 20 10 20 60 50 80 60.00%
1 100 1 100 202 203 302 48.77%
1 1000 1 1000 2002 2003 3002 49.88%
The gain percentage is affected by the magnitude of the numbers. The numbers above are a bad example and result from a primitive shift function.
You have to ask yourself which critter has the most productive gain:
Evaluate the growth of critters A, B, C, and D:
A used to consume 0.01 units of energy and now consumes 10 units.
B used to consume 500 units and now consumes 700 units.
C used to consume -50 units (Producing units!) and now consumes 30 units.
D used to consume -0.01 units (Producing) and now consumes -30 units (producing).
In some ways arguments can be made that each critter is the biggest grower in their own way. Some people say B is best grower, others will say D is a bigger gain. You have to decide for yourself which is better.
The question becomes, can we map this intuitive feel of what we label as growth into a continuous function that tells us what humans tend to regard as "awesome growth" vs "mediocre growth".
Growth a mysterious thing
You then have to take into account that Critter B may have had a far more difficult time than critter D. Critter D may have far more prospects for it in the future than the others. It had an advantage! How do you measure the opportunity, difficulty, velocity and acceleration of growth? To be able to predict the future, you need to have an intuitive feel for what constitutes a "major home run" and a "lame advance in productivity".
The first and second derivatives of a function will give you the "velocity of growth" and "acceleration of growth". Learn about those in calculus, they are super important.
Which is growing more? A critter that is accelerating its growth minute by minute, or a critter that is decelerating its growth? What about high and low velocity and high/low rate of change? What about the notion of exhausting opportunities for growth. Cost benefit analysis and ability/inability to capitalize on opportunity. What about adversarial systems (where your success comes from another person's failure) and zero sum games?
There is exponential growth, liner growth. And unsustainable growth. Cost benefit analysis and fitting a curve to the data. The world is far queerer than we can suppose. Plotting a perfect line to the data does not tell you which data point comes next because of the black swan effect. I suggest all humans listen to this lecture on growth, the University of Colorado At Boulder gave a fantastic talk on growth, what it is, what it isn't, and how humans completely misunderstand it. http://www.youtube.com/watch?v=u5iFESMAU58
Fit a line to the temperature of heated water, once you think you've fit a curve, a black swan happens, and the water boils. This effect happens all throughout our universe, and your primitive function (new-old)/old is not going to help you.
Here is Java code that accomplishes most of the above notions in a neat package that suits my needs:
Critter growth - (a critter can be "radio waves", "beetles", "oil temprature", "stock options", anything).
public double evaluate_critter_growth_return_a_gain_percentage(
double old_value, double new_value) throws Exception{
double abs_old = Math.abs(old_value);
double abs_new = Math.abs(new_value);
//This is your shift function, fool around with it and see how
//It changes. Have a full battery of unit tests though before you fiddle.
double biggest_absolute_value = (Math.max(abs_old, abs_new)+1)*2;
if (new_value <= 0 || old_value <= 0){
new_value = new_value + (biggest_absolute_value+1);
old_value = old_value + (biggest_absolute_value+1);
}
if (old_value == 0 || new_value == 0){
old_value+=1;
new_value+=1;
}
if (old_value <= 0)
throw new Exception("This should never happen.");
if (new_value <= 0)
throw new Exception("This should never happen.");
return (new_value - old_value) / old_value;
}
Result
It behaves kind-of sort-of like humans have an instinctual feel for critter growth. When our bank account goes from -9000 to -3000, we say that is better growth than when the account goes from 1000 to 2000.
1->2 (1.0) should be bigger than 1->1 (0.0)
1->2 (1.0) should be smaller than 1->4 (3.0)
0->1 (0.2) should be smaller than 1->3 (2.0)
-5-> -3 (0.25) should be smaller than -5->-1 (0.5)
-5->1 (0.75) should be smaller than -5->5 (1.25)
100->200 (1.0) should be the same as 10->20 (1.0)
-10->1 (0.84) should be smaller than -20->1 (0.91)
-10->10 (1.53) should be smaller than -20->20 (1.73)
-200->200 should not be in outer space (say more than 500%):(1.97)
handle edge case 1-> -4: (-0.41)
1-> -4: (-0.42) should be bigger than 1-> -9:(-0.45)
Simplest solution is the following:
=(NEW/OLD-1)*SIGN(OLD)
The SIGN() function will result in -1 if the value is negative and 1 if the value is positive. So multiplying by that will conditionally invert the result if the previous value is negative.
Percentage growth is not a meaningful measure when the base is less than 0 and the current figure is greater than 0:
Yr 1 Yr 2 % Change (abs val base)
-1 10 %1100
-10 10 %200
The above calc reveals the weakness in this measure- if the base year is negative and current is positive, result is N/A
It is true that this calculation does not make sense in a strict mathematical perspective, however if we are checking financial data it is still a useful metric. The formula could be the following:
if(lastyear>0,(thisyear/lastyear-1),((thisyear+abs(lastyear)/abs(lastyear))
let's verify the formula empirically with simple numbers:
thisyear=50 lastyear=25 growth=100% makes sense
thisyear=25 lastyear=50 growth=-50% makes sense
thisyear=-25 lastyear=25 growth=-200% makes sense
thisyear=50 lastyear=-25 growth=300% makes sense
thisyear=-50 lastyear=-25 growth=-100% makes sense
thisyear=-25 lastyear=-50 growth=50% makes sense
again, it might not be mathematically correct, but if you need meaningful numbers (maybe to plug them in graphs or other formulas) it's a good alternative to N/A, especially when using N/A could screw all subsequent calculations.
You should be getting a negative result - you are dividing by a negative number. If last year was negative, then you had negative growth. You can avoid this anomaly by dividing by Abs(Last Year)
Let me draw the scenario.
From: -303 To 183, what is the percentage change?
-303, -100% 0 183, 60.396% 303, 100%
|_________________ ||||||||||||||||||||||||________|
(183 - -303) / |-303| * 100 = 160.396%
Total Percent Change is approximately 160%
Note: No matter how negative the value is, it is treated as -100%.
The best way to solve this issue is using the formula to calculate a slope:
(y1-y2/x1-x2)
*define x1 as the first moment, so value will be "C4=1"
define x2 as the first moment, so value will be "C5=2"
In order to get the correct percentage growth we can follow this order:
=(((B4-B5)/(C4-C5))/ABS(B4))*100
Perfectly Works!
Simplest method is the one I would use.
=(ThisYear - LastYear)/(ABS(LastYear))
However it only works in certain situations. With certain values the results will be inverted.
It really does not make sense to shift both into the positive, if you want a growth value that is comparable with the normal growth as result of both positive numbers. If I want to see the growth of 2 positive numbers, I don't want the shifting.
It makes however sense to invert the growth for 2 negative numbers. -1 to -2 is mathematically a growth of 100%, but that feels as something positive, and in fact, the result is a decline.
So, I have following function, allowing to invert the growth for 2 negative numbers:
setGrowth(Quantity q1, Quantity q2, boolean fromPositiveBase) {
if (q1.getValue().equals(q2.getValue()))
setValue(0.0F);
else if (q1.getValue() <= 0 ^ q2.getValue() <= 0) // growth makes no sense
setNaN();
else if (q1.getValue() < 0 && q2.getValue() < 0) // both negative, option to invert
setValue((q2.getValue() - q1.getValue()) / ((fromPositiveBase? -1: 1) * q1.getValue()));
else // both positive
setValue((q2.getValue() - q1.getValue()) / q1.getValue());
}
These questions are answering the question of "how should I?" without considering the question "should I?" A change in the value of a variable that takes positive and negative values is fairly meaning less, statistically speaking. The suggestion to "shift" might work well for some variables (e.g. temperature which can be shifted to a kelvin scale or something to take care of the problem) but very poorly for others, where negativity has a precise implication for direction. For example net income or losses. Operating at a loss (negative income) has a precise meaning in this context, and moving from -50 to 30 is not in any way the same for this context as moving from 110 to 190, as a previous post suggests. These percentage changes should most likely be reported as "NA".
Just change the divider to an absolute number.i.e.
A B C D
1 25,000 50,000 75,000 200%
2 (25,000) 50,000 25,000 200%
The formula in D2 is: =(C2-A2)/ABS(A2) compare with the all positive row the result is the same (when the absolute base number is the same). Without the ABS in the formula the result will be -200%.
Franco
Use this code:
=IFERROR((This Year/Last Year)-1,IF(AND(D2=0,E2=0),0,1))
The first part of this code iferror gets rid of the N/A issues when there is a negative or a 0 value. It does this by looking at the values in e2 and d2 and makes sure they are not both 0. If they are both 0 then it will place a 0%. If only one of the cells are a 0 then it will place 100% or -100% depending on where the 0 value falls. The second part of this code (e2/d2)-1 is the same code as (this year - lastyear)/Last year
Please click here for example picture
I was fumbling for answers today, and think this would work...
=IF(C5=0, B5/1, IF(C5<0, (B5+ABS(C5)/1), IF(C5>0, (B5/C5)-1)))
C5 = Last Year, B5 = This Year
We have 3 IF statements in the cell.
IF Last Year is 0, then This Year divided by 1
IF Last Year is less than 0, then This Year + ABSolute value of Last Year divided by 1
IF Last Year is greater than 0, then This Year divided by Last Year minus 1
Use this formula:
=100% + (Year 2/Year 1)
The logic is that you recover 100% of the negative in year 1 (hence the initial 100%) plus any excess will be a ratio against year 1.
Short one:
=IF(D2>C2, ABS((D2-C2)/C2), -1*ABS((D2-C2)/C2))
or confusing one (my first attempt):
=IF(D2>C2, IF(C2>0, (D2-C2)/C2, (D2-C2)/ABS(C2)), IF(OR(D2>0,C2>0), (D2-C2)/C2, IF(AND(D2<0, C2<0), (D2-C2)/ABS(C2), 0)))
D2 is this year, C2 is last year.
Formula should be this one:
=(thisYear+IF(LastYear<0,ABS(LastYear),0))/ABS(LastYear)-100%
The IF value if < 0 is added to your Thisyear value to generate the real difference.
If > 0, the LastYear value is 0
Seems to work in different scenarios checked
This article offers a detailed explanation for why the (b - a)/ABS(a) formula makes sense. It is counter-intuitive at first, but once you play with the underlying arithmetic, it starts to make sense. As you get used to it eventually, it changes the way you look at percentages.
Aim is to get increase rate.
Idea is following:
At first calculate value of absolute increase.
Then value of absolute increase add to both, this and last year values. And then calculate increase rate, based on the new values.
For example:
LastYear | ThisYear | AbsoluteIncrease | LastYear01 | ThisYear01 | Rate
-10 | 20 | 30 = (10+20) | 20=(-10+30)| 50=(20+30) | 2.5=50/20
-20 | 20 | 40 = (20+20) | 20=(-20+40)| 60=(20+40) | 3=60/2
=(This Year - Last Year) / (ABS(Last Year))
This only works reliably if this year and last year are always positive numbers.
For example last_year=-50 this_year = -1. You get -100% growth when in fact the numbers have improved a great deal.
Suppose 5 samples of hue are taken using a simple HSV model for color, having values 355, 5, 5, 5, 5, all a hue of red and "next" to each other as far as perception is concerned. But the simple average is 75 which is far away from 0 or 360, close to a yellow-green.
What is a better way to calculate this mean and associated std?
The simple solution is to convert those angles to a set of vectors, from polar coordinates into cartesian coordinates.
Since you are working with colors, think of this as a conversion into the (a*,b*) plane. Then take the mean of those coordinates, and then revert back into polar form again. Done in matlab,
theta = [355,5,5,5,5];
x = cosd(theta); % cosine in terms of degrees
y = sind(theta); % sine with a degree argument
Now, take the mean of x and y, compute the angle, then
convert back from radians to degrees.
meanangle = atan2(mean(y),mean(x))*180/pi
meanangle =
3.0049
Of course, this solution is valid only for the mean angle. As you can see, it yields a consistent result with the mean of the angles directly, where I recognize that 355 degrees really wraps to -5 degrees.
mean([-5 5 5 5 5])
ans =
3
To compute the standard deviation, it is simplest to do it as
std([-5 5 5 5 5])
ans =
4.4721
Yes, that requires me to do the wrap explicitly.
I think the method proposed by user85109 is a good way to compute the mean, but not the standard deviation:
imagine to have three angles: 180, 180, 181
the mean would be correctly computed, as a number aproximately equal to 180
but from [180,180,-179] you would compute a high variance when in fact it is near zero
At first glance, I would compute separately the means and variances for the half positive angles , [0 to 180] and fot the negative ones [0,-180] and later I would compute the combined variance
https://www.emathzone.com/tutorials/basic-statistics/combined-variance.html
taking into account that the global mean and the difference between it and the local means has to be computed in both directions: clockwise and counterclockwise, and the the correct one has to be chosen.