stuck on a simple cumsum prob

I have a variable x. I make a variable y. I now need to regenerate my variable xNew from y.

Its very close but not exact. Why not? what I have done wrong?

x = cumsum(randn(1000,1));

y = 0.5.*(x(3:end) - x(1:end-2));

xNew = cumsum(y);

plot(x(3:end)); hold all; plot(xNew);

Answer 1

[Oleg Komarov]

y = 0.5.*(x(3:end) - x(1:end-2));

Is linear interpolation between point 1 and 3, then between 2 and 4 and so on.

You cannot recover x from the linearly interpolated y unless you are given x(1) and x(end).

EDIT

Suppose you have 5 points only (for simplicity), then:

y2* = (x3-x1)/2;
y3* = (x4-x2)/2;
y4* = (x5-x3)/2;

I assume you know the series of y, i.e. are numbers, but x is unobserved, then you can see that the system has 5 unknowns (x1,...x5) in three equations. Thus, you can have a unique solution only if you are given x1 and x5.

Now, if you are given the x series for comparison reasons, you can guess x1 and x5 then retrieve the rest of the series and compare the distance between xNew and the original one. In an iterative fashion you can then tweak the guess and comapre again the two series, if the distance has dimished then continue tweaking the guess in the same direction until you reach distance = 0.

EDIT#2 You don't need exactly x1 and x5 but any two xi.

[Matlab2010]

can you recover x(2:end-1)?

[Oleg Komarov]

No. See example above.

Answer 2

[Thomas]

x = cumsum(randn(1000,1));

%y = (x(2:end) - x(1:end-1));

y=[x(1); (x(2:end) - x(1:end-1))];

xNew = cumsum(y);

plot(x(2:end)); hold all; plot(xNew);

isequal(x,xNew)

[Oleg Komarov]

This problem is kinda different. There's no linear interpolation here.

[Matlab2010]

yes. but what about "(x(3:end) - x(1:end-2));"

[Thomas]

Hmm. then as oleg says you need, x(1) and x(end) for each value to interpolate..