[f,x]=ecdf(y) definationf of x

Hallo everyone,

Now i just do the statsic data analysis by ecdf function.

so i have a matrix, and i change this matrix in to a colum.

[f,x]=ecdf(y)

y represent this colum. then i get a colum of f and a colum of x, and plot (x, f),

i would like to know what it means for x, f is the cumulative function, also the probility.

how they caculate f and x based on the y

Thank so much

Jialin

Answer 1

[Torsten]

https://de.mathworks.com/help/stats/ecdf.html

There are six literature reference at the bottom of the page.

Usually, the minimum and maximum value of the y-data is determined. The interval [ymin,ymax] is divided in a certain number of subintervals (bins). According to their value, the y-data are put into the bin where they belong. Then the number of y-data in each bin is divided by the total number of y-data. Let us call these values bi (i=1,...,n = number of bins) (note that they sum to 1). Then it is set

B1 = b1

B2 = b1+b2

...

Bn-1 = b1 + b2 + ... + bn-1

Bn = 1

and the B1,B2,...,Bn are plotted as ecdf.

[JM]

Thanks so much. That is really help me a lot.

[Torsten]

I forgot to mention that the x are the midpoints of the bins and the y are the Bi.

[JM]

Thousands Thanks.

You are superman.

[JM]

hallo Torsten

I just want to confirm, this following diagram is from plot(x,f)

this x axis is minpoint of bins.right?

for this histogram, the data is also based on the y.

for this y axis, is also frequency. and this x axis, i don't understand why there is negative value, actually all the value from Y is positive.

[Torsten]

this x axis is minpoint of bins.right?

Yes, your y-data seem to range between 0 (ymin) and 0.8 (ymax).

i don't understand why there is negative value, actually all the value from Y is positive.

My guess is that you chose a normal distribution to approximate your empirical pdf - and the normal distribution has mass on the complete real axis (thus also for y being negative). That's why approximating your empirical pdf by a normal distribution is most probably a bad choice. Try a LogNormal distribution or something advised in the literature for your application.