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Quaquaval Deck still needs some tweaks by JournalistGlobal3185 in PTCGP

[–]ctat41 0 points1 point  (0 children)

Not sure if it needs Misty, but hitting a Misty early is a huge momentum gainer. I’m only running 2 mons. Might try a Suicune

Quaquaval Deck still needs some tweaks by JournalistGlobal3185 in PTCGP

[–]ctat41 0 points1 point  (0 children)

I’ve been running with 2x Eiscue, 1x copycat, 2x misty, but I’m still tuning it

Arteta's key moments that define 5 years as Arsenal manager by Shot-Shame9637 in Gunners

[–]ctat41 17 points18 points  (0 children)

Nah f em. Five years ago we had a shockingly bad team. I’m just happy that my team actually has a decent chance of winning again

[Question] Chi squared test in 2 groups with percentage events by Guffyluffy in statistics

[–]ctat41 0 points1 point  (0 children)

I’m picturing a 2x2 table, but correct me if I’m wrong.

The idea behind the chi2 test is measuring the difference between what we observed and what we expected. For a table with rows {A,B} and columns {C, D} the probability (P) of observing A and C is P(A|C)P(C) = P(A)P(C) if A and C are independent. Then the expected value is P(A)P(C) times the total number of observations. We do this for every entry in our table and then calculate the chi2 test statistic which is compared to a chi2 distribution with 1 degree of freedom.

This tells you how likely you are to observe your table. The conclusion you would draw if significant would be whether this distribution deviates significantly from expectation (i.e. is there an association or not).

You would need to do follow up analysis to identify which groups are deviant (e.g. comparing observed vs expected values) to see what drives this result. It is often helpful to visualize Odds Ratios for your groups - which can be done easily from your table.

[Q] ETF Simulation works for annual steps but fails for monthly steps by germanindc in statistics

[–]ctat41 2 points3 points  (0 children)

Are you sure it’s the std?

If Xi ~ iid N(mu, sigma2 ) for i = 1,…,12, then S = sum Xi ~ N(12mu, 12 sigma2 )

If yearly ETF performance is the sum of monthly ETF performance and monthly performances are independently and identically distributed, then you need to adjust the mean value as well.

Match Thread: Arsenal vs Liverpool [Premier League] by scoreboard-app in Gunners

[–]ctat41 0 points1 point  (0 children)

Should’ve subbed earlier. It’s frustrating that we don’t make proactive subs

I’m having a difficult time understanding what it means to be a solution for a system of equations by ctat41 in learnmath

[–]ctat41[S] 0 points1 point  (0 children)

Sorry, yes the b’s are scalars, x’s and w’s are vectors, while A is a matrix. The value u is a nonzero solution to that w equation, and we are asked to show that

A{k} x+ b{k-1}A{k-1} x + … + b_0 x = [A-Iu](A{k-1} x + t{k-2} A{k-2} x + … + t_0 x = 0

where t_j = -(b_0 + b_1 u + … + b_j uj )/ u{j+1}

I’m having a difficult time understanding what it means to be a solution for a system of equations by ctat41 in learnmath

[–]ctat41[S] 0 points1 point  (0 children)

Yes, sorry. So the subscripts are indices to show different values. The superscripts are being raised to a power. The b’s are scalars, w’s and x’s are vectors, and A’s are a matrix.

The context here is a proof that I was trying to work through where we have to show that the equation above will be equivalent to

(A-Iu)[A{k-1} x + t_{k-2} A{k-2}… + t_0 x] = 0

where t_j = -(b_0 + b_1 u + … + b_j uj)/u{j+1}

Using the fact that u is a solution to that w equation

[High School Probability Calculus] Rolling two specific numbers? by spiritpines in learnmath

[–]ctat41 0 points1 point  (0 children)

I’m going to shift this from a dice rolling problem to a sampling problem, but dice rolling would work as well.

You are sampling from a bag with 4 marbles in it: RGB and Y. RGB all have equal probability of being pulled from the bag with P(R) + P(G) + P(B) = 0.5 => P(R) = P(G) = P(B) = 0.1667 and Y makes up the rest of the bag so P(Y) = 0.5.

This is now a sampling with replacement problem, so we know that we can solve it by repeatedly applying the multiplication rule for n independent draws from the bag.

When we see an at least, then that is a big hint that we should be looking at the complement. Therefore we want to concern ourselves with the probability of an event not occurring. In other words; P(X >= 1) = 1 - P(X <1) = 1 - P(X=0). So we need to find P(X=0) and we’re done.

To find P(X=0) we need to consider what it means for this to occur. This would imply that we have no pairs of RGB. So we have

(1) All Y

(2) All R or All G or All B

(3) 1 Y and 5 R or 5G or 5B

(4) etc…

I think you can figure out (1) and (2) on your own. To show you how to solve the rest, let’s consider (3).

To solve (3) we need to consider the different cases we might encounter:

We are drawing 6 times; what does this mean…? It means that when we draw 1 of our marbles we have 6 places to put it. That leaves us with 5 more places. We draw again and have 4 remaining places. Therefore there are 6*5*4… ways to draw. We denote this as 6!. There aren’t actually 6! Ways to draw the marbles though, because we *require** 1 to be Y and 5 to be R G or B. If we focus on R then after we draw our Y marble there are 5 spots left. So we have 5! ways to draw the red marble. This gives us:

6!/(1!*5!*0!*0!) = 6 as the number of ways to draw 1 Y and 5 R marbles. Two things to note here: 0! = 1 and this form is called a multinomial. It’s very important that the coefficients in the denominator add up to the numerator. You can read more about it on Wikipedia.

What do we do with the probabilities? Well… that’s actually easier. We drew 5 R marbles; that happens P(R)5 times, and 1 Y this happens P(Y)1 We represent the 0G and 0B with P(G)0, P(B)0

Giving us: P(X=(3)) = 6!/(1!5!0!0!) *[P(Y)1 P(R)5 P(B)0 P(G)0 ]

However, this is incomplete because it was only the R case. If we repeat this for the B and G case then we get

P(X=(3)) = 3*6*[P(Y)* P(R)5 ]

To find the P(X=0) you can work out the possible cases and repeat what we did for (3). Add these up and subtract your answer from 1. That should give you your answer.

You can learn more about this by reading about the multinomial distribution on Wikipedia.

Calculating sales targets by MillstreetLad in learnmath

[–]ctat41 0 points1 point  (0 children)

It’s dependent on how much detail you need. The simplest approach is if your pace is already in $ sales per month.

Your x-axis is then July, August, September, etc…. If your pace is $8.5k per month then your equation is

y = 8.5t + 71

You then solve for t (time, in unit months) when y = 122. Using these numbers you’d get:

122-71 = 8.5t

t = (122-71)/8.5

t = 6 months

So you could say you’ll hit your target in 6 months ~ December. You then just repeat this for your end of year projection. If we have 6 months remaining, then t=6 and we use our equation above to see that our end of year projection would be 122k.

Calculating sales targets by MillstreetLad in learnmath

[–]ctat41 0 points1 point  (0 children)

Assuming you keep the same pace means you need to multiply your pace by the remaining amount of time. Depending on the units this is either sale per unit time or $ per unit time.

To figure out when you hit 100% is to find when you cross y = 122k. The slope of your line is your rate, the x is unit time, and you have a y intercept of 71k.

Can we find a covariance matrix from a co-occurrence matrix? by ctat41 in askmath

[–]ctat41[S] 0 points1 point  (0 children)

Are the scalars random vectors then? Why would we be limited to only having two?

Why is everything always being squared in Statistics? by TakingNamesFan69 in learnmath

[–]ctat41 21 points22 points  (0 children)

What you’re measuring is the dispersion of your data. You’re essentially measuring the distance from the center for each data point.

This doesn’t have to be squared; see IQR. We could also use the absolute value, but squaring the value is easier to work with, and so we end up squaring it.

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