Here we show how the coefficients from the quadnomial probability distribution fit the multinomial formula. Due to symmetry of a quadnomial vector taken to the fourth power, the number of matrix multiplications is minimized to the square of two identical 4x4 diagonal symmetrical matrices.
We start with the total probability of four variables, which we only need to take to no greater than the
same power as the number of variables (ie., four for a quadnomial).
w + x + y + z
= 1
(w + x + y + z)1
= w
+ x + y + z
(w + x + y + z)2
= |w ||w
x y z |
|x |
|y |
|z |
|x |
|y |
|z |
= |w2
wx wy wz |
|xw
x2 xy xz |
|yw
yx y2 yz |
|zw
zx zy z2 |
(w + x + y + z)3
= |w2
wx wy wz ||w
x y z |
|xw
x2 xy xz |
|yw
yx y2 yz |
|zw
zx zy z2 |
We are able to skip here from three trials (e.g., third power), to completely solving for four trials (e.g., fourth power) of the the quadnomial.
(w + x + y + z)4
= [(w
+ x + y + z)2]2
= |w2
wx wy wz ||w2
wx wy wz |
|xw
x2 xy xz ||xw
x2 xy xz |
|yw
yx y2 yz ||yw
yx y2 yz |
|zw
zx zy z2 ||zw
zx zy z2 |
= |a11
a12 a13 a14|
|a21
a22 a23 a24|
|a31
a32 a33 a34|
|a41
a42 a43 a44|
As an example, let’s solve for |a21|, which of course also equals |a12| due to diagonal symmetry.
|a21|
= |w2 || |
|xw
||xw
x2 xy xz |
|yw
|| |
|zw
|| |
= |w3x
w2x2 w2xy w2xz |
|w2x2
wx3 wx2y wx2z |
|w2xy
wx2y wxy2 wxyz |
|w2xz
wx2z wxyz wxz2 |
Now we collect all the coefficients from all of the |abc| matrices, including |a21|. The coefficients for all of the variable combinations are:
1 w4
4 w3x
4 w3y
4 w3z
1 x4
4 wx3
4 x3y
4 x3z
1 y4
4 wy3
4 xy3
4 y3z
1 z4
4 wz3
4 xz3
4 yz3
12 w2xy
12 w2xz
12 w2yz
6 w2x2
6 w2y2
6 w2z2
6 x2y2
6 x2z2
6 y2z2
12 wx2y
12 wx2z
12 x2yz
12 wy2x
12 wy2z
12 xy2z
12 wxz2
12 wyz2
12 xyz2
24 wxyz
256 sum
of coefficients
= 44 or
number of variables to the power the vector is taken to (in this case also 4)
We can see from above that there are five classes of variable combinations and the coefficients equal the combination formula for multinomials, or 4P#w,#x,#y,#z = 4!/(#w's! #x's! #y's! #z's!):
()4 or w4, y4, y4, z4 = 4!/(4!) = 1
()3() = 4!/(3!) = 4
()2()2 = 4!/(2!2!) = 6
()2()() = 4!/(2!1!1!) = 12
()()()() or wxyz = 4!/(1!1!1!1!) = 24
So these quadnomial coefficients values align to the purple list above of the 35 variable combinations. And as an easy test, if all variables have a 25% chance of occurring, then the total probability is equal to 256(25%)4 = 256/44 = 1.
No comments:
Post a Comment