3 7 Example Bioassay experiment Problem statement Observations

3. 7 Example: Bioassay experiment • Problem statement – Observations: At each level of dose, 5 animals are tested, and number of death are observed. As dose x is increased, # of death increases. – Objective: estimate unknown probability of death , given dose x. Note that is not a single parameter but varies w. r. t. x. -1 -

Establishing posterior distribution • Binomial distribution – Let the unknown death probability be i for xi for ith group. – Then the data yi are binomially distributed: • Modeling dose response relation – is not constant but a function of dose x. Simplest trial is linear function. – This model has flaw that for very low or high x, it approaches ±∞. To meet this condition, introduce logistic transformation Then leads to – Relation is changed to This is called a logistic regression model. -2 -

Establishing posterior distribution • The likelihood – Likelihood of each individual group – Remark: model is characterized by two parameters , , not . Hence, these are the parameters to be estimated. • Joint posterior pdf – Prior distribution is assumed uniform or non-informative, i. e. , – Remark: can we write down the expression in closed form ? Never seen the pdf like this, nor are there standard pdfs available. Nevertheless, this is probability distribution indeed. -3 -

Analyzing posterior distribution • Rough estimate of the parameters – We can crudely estimate these by least squares minimization (linear regression) of zi on xi where i=1, …, 4. The solution for ( , ) are (0. 1, 2. 9). – We can also estimate standard error of ( , ) The solution for (V , V ) are (0. 3, 0. 5). – This gives us rough range of the parameters. ~ 0. 1 ± 3*0. 3 and ~ 2. 9 ± 3*0. 5 -4 -

Analyzing posterior distribution • Contour or 3 -D plot of joint posterior density – Define the domain as mean ± 3 std errors, which are [0. 1 0. 9]=[-1, 1] x [2. 9 1. 5]=[1, 5]. – Use computer program to plot contour, and find out the range is wrong indicating that the crude estimation was really crude indeed. – After trial & error, find out much wider range is needed, [-5, 10]x[-10, 40]. • Remarks – How can we analyze this distribution ? i. e. , how can we get the mean, variance, confidence intervals, etc ? -5 -

Sampling from posterior distribution • Grid method (inverse cdf method) – Refer section 1. 9 of Gelman or 9. 2. 3 of Mahadevan. • Procedure – 1. 2. 3. In order to generate samples following pdf f(v), Construct cdf F(v) which is the integral of f(v). Draw random value U from the uniform distribution on [0, 1]. let v=F-1(U). Then the value v will be a random draw from f(v). • Practice – Generate samples for N(0. 1). – Validate with analytic solution by comparing pdf shape mean, std, 95% conf bounds. -6 -

Sampling from posterior distribution • Factorization approach – Sample the two parameters & in the same way as the two parameters of normal distribution. where p( |y): marginal pdf of which is obtained by integration. p( | , y): conditional pdf of b given . • Procedure 1. 2. 3. 4. 5. Normalize the discrete joint pdf, i. e. , make the total probability value 1. Make marginal pdf p( |y) by simply summing over . Draw from marginal pdf of using grid method. Draw from conditional pdf of |. Post-analyze using the samples: As we have obtained samples of & , one can evaluate several characteristics. -7 -

Sampling from posterior distribution • Sample results – Scatter plot & comparison with contour plot – Posterior distribution of LD 50, which is the dose level x at which the prob. death is 0. 5. Simulating this is trivial. -8 -

Posterior prediction by sampling • Original definition – In this problem, probability of death is predicted at a dose level x. • Practical solution – For each sample a & b, generate new y (which is prob. death q here). – By this way, simulation is much more convenient than the analytic (or numerical) integration. -9 -

Remarks on grid based method • Remarks – There can be difficulty finding correct location & scale for the grid points. If the area is too small, it can miss the important area. If the area is too large, it can miss the importance due to large interval. – When computing posterior pdf, overflow / underflow encountered usually because it is multiplication of individual pdfs. To avoid, introduce log to the posterior density. – Computation grow prohibitly as more parameters are included. 500 grid for a parameter leads to 500^4 = 6. 25 e 10 numbers for the whole grid, at which the function value should be evaluated. Conclusion: this method is not used in practice. - 10 -