OUTLINE OF A PROOF OF FISHER'S THEOREM

The p linear forms  Y_i (  i = 1, ..., p) generate a p-dimensional hyperplane spanned by p orthogonal axes Y₁, ..., Y_p. The supplementary subspace Y is (n - p)-dimensional, and is easily fitted with (n - p ) additional orthogonal axes Y_p+1, ..., Y_n, each axis being also orthogonal to each of the Y_i (  i = 1, ..., p).

Each of the Y_i (i = p+1, ..., n) carries a variable Y_i that is  ~N(0, 1) (owing to the spherical symmetry of the joint distribution of the X_i), and these Y_i are independent (owing to the orthogonality of the Y_i).

In the lower image:

    * p = 1, and Y₁ is the axis defined by the linear form Y₁.

    * The supplementary space is the red plane spanned by Y₂ and Y₃.

    * X_P is the projection of X on this plane.

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Let's now get back to the interpretation of Q:

Q(X₁, X₂ , ..., X_n) = (_iX_i ²)  - Y₁² - Y₂² -...- Y_p²             p < n  

This interpretation has first a geometric part, then a probabilistic part.
 

Geometry

    1) _iX_i ²  is just the square of the distance of X = (X₁, ..., X_n) to the origin O.

    2) From this quantity, we subtract the squares of the projections of X on p orthogonal axes (the Y_i,   i = 1, ..., p).

    3) We then create n - p supplementary axes as stated above.

    4) The Pythagorean theorem then states that what's left (i.e. Q) is the sum of the squares of the projections of X on these supplementary axes. In fact, it states that Q is the square of the distance of the origin to X_P, the projection of X on the supplementary subspace.

So the square of the length of OX_P is:

OX_P² =_iY_i ²          i = p+1, ..., n

Probabilities

    * Each of the Y_i is N(0, 1) and the Y_i are independent (see above). So the square of the length of OX_P is the sum of n - p squared independent variables (on the figure, Y₂ and Y₃) all N(0, 1), and is therefore distributed as _{n - p} .

    * Besides, Q is a function of variables (the supplementary Y_i) that are all independent of the Y_i,   i = 1, ..., p, and is therefore independent of any of them.