Probability assessment is purely mathematical - page 5
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And a problem that has been plaguing me for a long time.
The balance is conventionally 0. Wandering in minus and plus at random with no spread.
How many times should I expect to have balance state=0, with 100 iterations?
Kolmogorov "Introduction to Probability Theory" https://www.mql5.com/go?link=http://www.mirknig.com/knigi/1181165246-vvedenie-v-teoriju-verojatnostejj.html pp. 88-89
Thanks, I can't download it yet. But I will definitely have a look.
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Another question then!
Is it possible to draw a line (minimum sum of quadratic deviations) - the usual solution is on the forum... but!
For example to draw such a line for Сlose - with a condition that the line must necessarily pass through, Close[0]?
int start()
{ int limit;
int counted_bars=IndicatorCounted();
//---- last counted bar will be recounted
if( counted_bars>0) counted_bars--;
limit=Bars-counted_bars;
double a,b,c, sumy, sumx, sumxy, sumx2;
for(int j=limit; j>=0; j--)
{ sumy=0=0; sumxy=0 0; sumx2=0 0; for(int i=0; i< barsToCount; i++99 {0;
sumx=0.0;
sumxy=0.0;
sumx2=0.0;
for(int i=0; i< barsToCount; i++)
{
sumy+=Close[i+j];
sumxy+=Close[i+j]*i;
sumx+=i;
sumx2+=i*i;
}
c=sumx2*barsToCount-sumx*sumx;
if(c==0.0)
{
Alert("LinearRegression error: can\'t resolve equation");
return;
}
b=( sumxy*barsToCount-sumx*sumy)/c;
a=( sumy-sumx*b)/barsToCount;
bufferB[j]=a;
bufferE[j]=a+b*barsToCount;
}
This is a classic.
If you "weight" Close[0], it would probably have the right effect.
But how?
Since the topic of probabilities came up, I wanted to ask a question.
We have two unobservable events, which with some probability (each event has its own probability) "trigger" the same process. How do we calculate the probability of both of these events happening at the same time?
For example, if a dry tree branch breaks with probability 0.6. If a squirrel sits on the branch, the probability is 0.3. What if it is a dry tree and a squirrel is sitting? It's all about the average. But it doesn't make sense. It turns out that if we remove the squirrel, the probability increases :)
A school question, but I'm confused :(
0.6 * 0.3 = 0.18
wrong
1-0.4*0.7 = 0.72
wrong
1-0.4*0.7 = 0.72
Right! Thank you.
To explain.
The probability that a branch will not break when the factors act simultaneously is (1-0.6)*(1-0.3)