From: Wang, Yaning <*yaning.wang*>

Date: Wed, 18 Feb 2009 13:05:44 -0500

Bihorel:

If you are familiar with the concept of hazard in survival analysis, this conditional likelihood should be straightforward. A slight modification of h(y)=f(y)/S(y) to h(y)=f(y)/S(LOQ) where (y>=LOQ) will give you the expression used by M2: l(y)=f(y)/S(LOQ) at a specific time t. You should ignore t in the M2 expression and think of it as l(y)=P(Y=y|Y>=LOQ) where y>=LOQ at time t. P(Y=y|Y>=LOQ) is not a discrete probability but a probability density because y is a continuous variable.

I think P(Y=y, Y>=LOQ) can be simplified to P(Y=y) because there is an inherent restriction: y>=LOQ.

Yaning

Yaning Wang, Ph.D.

Team Leader, Pharmacometrics

Office of Clinical Pharmacology

Office of Translational Science

Center for Drug Evaluation and Research

U.S. Food and Drug Administration

Phone: 301-796-1624

Email: yaning.wang

"The contents of this message are mine personally and do not necessarily reflect any position of the Government or the Food and Drug Administration."

-----Original Message-----

From: owner-nmusers

Sent: Monday, February 16, 2009 5:22 PM

To: Leonid Gibiansky

Cc: nmusers

Subject: Re: [NMusers] Theoretical questions about Beal's M2 method

Thanks Leonid,

However, there is still one point that is unclear to me. You have

demonstrated that p(y | y > LLQ) = p(y) / p(y>LOQ), given the

assumptions of the text. Now, this is a discrete probability, while l(t)

is a likelihood... How can one mathematically demonstrate the expression

of l(t) used by Dr. Beal starting from the previous expression of p(y |

y > LLQ)?

*Sebastien Bihorel, PharmD, PhD*

PKPD Scientist

Cognigen Corp

Email: sebastien.bihorel

<mailto:sebastien.bihorel

Phone: (716) 633-3463 ext. 323

Leonid Gibiansky wrote:

*> You can view it as:
*

*>
*

*> p(y $B"A(J y > LLQ) = 0 when y < LLQ
*

*> p(y $B"A(J y > LLQ) = p(y) when y > LLQ
*

*>
*

*> Another way to look on this is to say that
*

*> p(y | y > LLQ) is proportional to p(y) and should integrate to 1
*

*>
*

*> integral(p(y)) over y > LLQ is ( 1- phi((LLQ-f(t)/g(t))) that
*

*> immediately leads to l(t) below.
*

*>
*

*> As to the 0 to 1 restriction, l(t) is the density, not probability. It
*

*> should integrate to one but can be smaller or greater than 1 (any
*

*> positive number).
*

*>
*

*> Leonid
*

*>
*

*>
*

*>
*

*> --------------------------------------
*

*> Leonid Gibiansky, Ph.D.
*

*> President, QuantPharm LLC
*

*> web: www.quantpharm.com
*

*> e-mail: LGibiansky at quantpharm.com
*

*> tel: (301) 767 5566
*

*>
*

*>
*

*>
*

*>
*

*> Sebastien Bihorel wrote:
*

*>> Dear colleagues,
*

*>>
*

*>> In a paper dated from 2001, Dr. Beal presented several methods to
*

*>> handle data below the quantification limit (Journal of
*

*>> Pharmacokinetics and Pharmacodynamics, Vol. 28, No. 5, October 2001),
*

*>> including the M2 method that can be implemented in NONMEM 6 via the
*

*>> YLO functionnality. I would like to submit some questions to the list
*

*>> about the theory associated to the M2 method.
*

*>>
*

*>> I quote:
*

*>> "...the BQL observations can be discarded, and under the assumption
*

*>> that all the D(t) [the distribution of residual errors] are normal,
*

*>> the method of maximum conditional likelihood estimation can be
*

*>> applied to the remaining observations (method M2). With this method,
*

*>> the likelihood for the data, conditional on the fact that by design,
*

*>> all (remaining) observations are above the QL, is maximized with
*

*>> respect to the model parameters. The density function of the
*

*>> distribution on possible observations at time t, evaluated at y(t),
*

*>> is 1/sqrt( 2*pi*g(t) ))*exp( -0.5*( y(t)-f(t) )^2/g(t) ) and the
*

*>> probability that an observation at time t is above the QL is 1-
*

*>> phi((QL-f(t)/g(t))), where phi is the cumulative normal distribution
*

*>> function. Therefore, conditional on the observation at time t being
*

*>> above QL, the likelihood for y(t) is the ratio:
*

*>> l(t)=(1/sqrt( 2*pi*g(t) ))*exp( -0.5*( y(t)-f(t) )^2/g(t) ) /( 1-
*

*>> phi((QL-f(t)/g(t))) [equation 1]"
*

*>>
*

*>> Now, lets A and B be two events. The probability of A, given B is:
*

*>> p(A|B) = p(A$B"A(JB) / p(B)
*

*>>
*

*>> In the context of Dr. Beal's paper, I interpret A as simply the
*

*>> observation y(t) and B as the fact that y(t) is above QL, and thus
*

*>> have the following questions about equation 1:
*

*>> - it looks like p(A$B"A(JB) in equation 1 simplifies to the probability of
*

*>> y(t) given the model parameters, i.e. p(A). Which part of the problem
*

*>> allows this simplification?
*

*>> - how can l(t) be constrained between 0 and 1 if both numerator and
*

*>> denominator can vary between 0 and 1?
*

*>>
*

*>> Any comment from nmusers will be greatly appreciated.
*

*>>*

Received on Wed Feb 18 2009 - 13:05:44 EST

Date: Wed, 18 Feb 2009 13:05:44 -0500

Bihorel:

If you are familiar with the concept of hazard in survival analysis, this conditional likelihood should be straightforward. A slight modification of h(y)=f(y)/S(y) to h(y)=f(y)/S(LOQ) where (y>=LOQ) will give you the expression used by M2: l(y)=f(y)/S(LOQ) at a specific time t. You should ignore t in the M2 expression and think of it as l(y)=P(Y=y|Y>=LOQ) where y>=LOQ at time t. P(Y=y|Y>=LOQ) is not a discrete probability but a probability density because y is a continuous variable.

I think P(Y=y, Y>=LOQ) can be simplified to P(Y=y) because there is an inherent restriction: y>=LOQ.

Yaning

Yaning Wang, Ph.D.

Team Leader, Pharmacometrics

Office of Clinical Pharmacology

Office of Translational Science

Center for Drug Evaluation and Research

U.S. Food and Drug Administration

Phone: 301-796-1624

Email: yaning.wang

"The contents of this message are mine personally and do not necessarily reflect any position of the Government or the Food and Drug Administration."

-----Original Message-----

From: owner-nmusers

Sent: Monday, February 16, 2009 5:22 PM

To: Leonid Gibiansky

Cc: nmusers

Subject: Re: [NMusers] Theoretical questions about Beal's M2 method

Thanks Leonid,

However, there is still one point that is unclear to me. You have

demonstrated that p(y | y > LLQ) = p(y) / p(y>LOQ), given the

assumptions of the text. Now, this is a discrete probability, while l(t)

is a likelihood... How can one mathematically demonstrate the expression

of l(t) used by Dr. Beal starting from the previous expression of p(y |

y > LLQ)?

*Sebastien Bihorel, PharmD, PhD*

PKPD Scientist

Cognigen Corp

Email: sebastien.bihorel

<mailto:sebastien.bihorel

Phone: (716) 633-3463 ext. 323

Leonid Gibiansky wrote:

Received on Wed Feb 18 2009 - 13:05:44 EST