Introduction to LOD Score Concept
Let QA = the probability
of success when HA is true; n = the total number
of trials; x = the number of successful events out of
n trials; P equal the probability of the observed results
when QA is the
specified value, using the binomial distribution. We can
use a LOD score to determine how much more likely it is
that the hypothesized probability of success (QA)
is true compared to the probability that Ho: Q
= ½. We first determine the probability of the observed
number of successful events occurring out of N trials
when QA is the
value specified by the alternative hypothesis. Next we
determine the probability of the observed number of successful
events occurring out of N trials when Q
= ½. We take the Log10 of the ratio of these
probabilities to determine the LOD score. A LOD score
of 3 means that the probability that HA is
true is 103 = 1000 times more likely than it
is that Q = ½.
The LOD score is the Log10 of the ratio of
the probability that the hypothesized probability of success
is QA versus the
probability that Q = ½.
Example:
An F2 family consists of 12 progeny, 8 progeny have the
genotype A_ while 4 progeny have the genotype aa. We will
use the LOD score to determine how much more likely it
is that QA = 3/4
versus Q = ½. Ho: Q=
½; HA: Q = 3/4
.
| P=L(QA/x) = |
n! |
Qx(1-Q)n-x |
| x!(n-x)! |
| = |
12! |
(3/4)8(1/4)4=0.193578(whenQ=3/4) |
8! 4! |
| P=L(Qn/x)= |
n! |
Qx(1-Q)n-x |
| x!(n-x)! |
| = |
12! |
(1/28(1/2)4 = 0.1205(whenQ = 1/2) |
| 8! 4! |
| LOD |
= Log10 |
L(QA/X) |
= Log10 |
0.193578 |
|
| L(Qn/X) |
0.1205 |
This means that the LOD is 0.204 or that the probability
that Q = 3/4
is 100.204 = 1.6 times more likely than the
probability that Q = 0.5
when we have a family size of 12 with 8 A_ progeny.
Calculation of the LOD score can be simplified, due to
the fact that the factor:
is a constant for the numerator and denominator of the
LOD score.
Then we can use:
| LOD |
= Log10 |
L(QA/X) |
|
| L(Qn/X) |
| |
= xLog(QA) + (n - x)Log(1 - QA) - nLog(0.5) |
In our example QA = 3/4 and
Qn = ½; x = 8; n = 12.
LOD = 8Log(3/4) + (12 - 8)Log (1 - 3/4) - 12 Log(½)
= 8(-0.1249387) + 4(-0.60206) - 12(-0.30103) = 0.204
A LOD of 0.204 means that it is 100.204 = 1.6
times more likely that Q = 3/4 than it is that Q
= 1/2.
The reason we can use the shortcut formula:
| LOD |
= Log10 |
L(QA/X) |
|
| L(Qn/X) |
| |
= xLog(QA) + (n - x)Log(1 - QA) - nLog(0.5) |
| |
is as follows:
| Log |
n! |
Qx(1-Q)n-x |
|
|
| x!(n-x)! |
|
| |
= Log |
n! |
+ xLog(Q)+(n-x)Log(1-Q) |
| x!(n-x)! |
| Log |
L(QA/X) |
=Log{L(QA/X)} - Log{L(Qn/X)} |
| L(Qn/X) |
| =Log |
n! |
+ xLog(QA) + (n - x)Log(1 - QA) |
| x!(n - x)! |
| -Log |
n! |
- xLog(Qn)) - (n - x)Log(1 - Qn) |
| x!(n - x)! |
= xLog(QA) + (n
- x)Log(1 - QA)
- xLog(Qn) - (n
- x)Log(1 - Qn)
When Qn = ½, then
the above equation can be further reduced to:
| Log |
L(QA/X) |
= xLog(QA) + (n - x)Log(1 - QA) - nLog(1/2) |
| L(Qn/X) |
because
| L(Qn/X) = |
n! |
(1/2)x(1 - 1/2)n - x |
| x!(n - x)! |
| = |
n! |
(1/2)x(1/2)n - x= |
n! |
(1/2)n |
| x!(n - x)! |
x!(n - x)! |
and the Log
| {L(Qn/X)} = Log |
n! |
+ nLog(1/2) |
| x!(n - x)! |
In
the case of linkage we have Ho: p = 0.5 versus HA: p =
QA. We can determine the LOD score at various values
of QA and when the LOD is at a peak, we have the value
of p that best fits the observed data.
LOD Score
LOD is an abbreviation for "log of the odds."
The lod score
| z = log10 |
[ |
L(QA|X) |
] |
| L(QN|X) |
Example, Binomial distribution lod score:
Let Q = probability(success);
1-Q = probability(failure);
Let x = number of successes;
n-x = number of failures.
The probability of X successes out of n trials is:
| n! |
QX(1-Q)(n-X) |
| x! (n-x)! |
Ho: Q = 0.5
HA: Q is unequal
to 0.5
| z = log10 |
[ |
L(QA|X) |
] |
| L(QN|X) |
z = xLog10
Q + (n-x)Log10
(1-Q) - nLog10(1/2)
This is because:
| = L(QA|x) = |
n! |
Qx
(1-Q)(n-x) |
| x!(n-x)! |
and
Log[L(QA|x)]
| = Log |
[ |
n! |
] |
+ xLog(Q) + (n - x)Log(1 - Q) |
| x! (n - x)! |
and
| = Log |
[ |
n! |
(0.5)x (0.5)n-x |
] |
| x! (n - x)! |
| = Log |
[ |
n! |
(0.5)n |
] |
| x! (n - x)! |
| = Log |
n! |
+ nLog(0.5) |
| x! (n - x)! |
Then
| z = Log10 |
[ |
L(QA|X) |
] |
| L(QN|X) |
| Log |
[ |
n! |
] |
+ xLog(Q) + (n - x)Log(1 - Q) |
| x! (n - x)! |
| - Log |
[ |
n! |
] |
- nLog(0.5) |
| x! (n - x)! |
z = xLog10(Q)
+ (n-x)Log10(1-Q)
- nLog10(0.5)
** Stated another way, the probability of observing
a sample of size n with x successes and n-x failures
when the probability of success in any one event equals
Q.
L(Q) = Probability(x successes
out of n trials with Q =
the probability of one successful event)
| = |
n! |
Qx (1-Q)(n-x) |
| x! (n - x)! |
Then when Q = 1/2, then
| L(Q) = |
n! |
(1/2)n(1/2)(n - x) |
| x! (n - x)! |
* L(QN) is not
the maximum likelihood estimate, it is the probability
of observing X successes out of n trials for a given
value of Q. L(QN)is
the maximum likelihood estimate of observing X successes
out of n trials when Q =
1/2.
