Description
Problem Six Neurons Ball Bearings Mating Calls
Total Score /33 /33 /34 /100
1. Six Neurons. Six neurons N1, N2, …, N6 are connected as in the right panel in Fig. 1.
If a stimulus is present, the neuron will fire with probability 0.9. When the stimulus is not
present, the neuron may still fire but with a small probability of 0.05. Firing of a neuron
serves as a stimulus for the subsequent neuron. N1 is given a stimulus.
Figure 1: Neuron Fires!
(a) What is the probability that N6 will fire?
(b) What is the probability that N6 will fire if N4 did not fire?
(c) If N6 did not fire, what is the probability that N5 received stimulus.
Hint: You can solve this problem by any of the 3 ways: (i) use of WinBUGS or OpenBUGS, (ii) direct simulation using Octave/MATLAB, R, or Python, and (iii) exact calculation. Use just one of the three ways to solve it.
2. Endurance of Deep Groove Ball Bearings. The data analyzed by Lawless (1986;
page 228)1 arose in tests on endurance of deep groove ball bearings (Fig. 2).
The data, in
units of 107
revolutions before failure for each of the 23 ball bearings in the life test, are:
1.788, 2.892, 3.300, 4.152, 4.212, 4.560, 4.880, 5.184, 5.196, 5.412, 5.556, 6.780, 6.864, 6.864,
6.888, 8.412, 9.312, 9.864, 10.512, 10.584, 12.792, 12.804, and 17.340.
Figure 2: Deep groove ball bearing
Assume that observations are coming from gamma Ga(r, λ) distribution, where shape
parameter is known, r = 4, and rate parameter λ is to be estimated in a Bayesian fashion.
1Lawless, J. F., 1982. Statistical Models and Methods for Lifetime Data, Wiley, New York.
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An expert elicits a gamma prior on λ,
π(λ) = λ
α−1β
α
Γ(α)
e
−βλ
,
with hyperparameters α = 3 and β = 5.
(a) Find Bayes estimator of λ (posterior mean), the 95% equitailed credible set for λ,
and the posterior probability of hypothesis H : λ ≤ 0.5.
Hint: No WinBUGS should be used, the problem is conjugate. You will need Octave or R
or Python, to calculate gamma cdf and quantiles.
Mating Calls. In a study of mating calls in the gray tree frogs Hyla hrysoscelis and Hyla
versicolor, Gerhart (1994)2
reports that in a location in Lousiana the following data on the
length of male advertisement calls have been collected:
Sample Average SD of
size duration duration
Hyla chrysoscelis 43 0.65 0.18
Hyla versicolor 12 0.54 0.14
The two species cannot be distinguished by external morphology, but H. chrysoscelis (Fig.
3) are diploids while H. versicolor are tetraploids. The triploid crosses exhibit high mortality
in larval stages, and if they attain sexual maturity, they are sterile. Females responding to
the mating calls try to avoid mismatches.
Figure 3: Hyla chrysoscelis
Assume that duration observations are normally distributed with means µ1 and µ2, and
precisions τ1 and τ2, for the two species respectively. For i = 1, 2, assume normal priors on
µi
’s as N (0.6, 1) and gamma priors on τi
’s as Ga(20, 0.5), where 0.5 is a rate hyperparameter.
Do the following simulations in Octave (MATLAB), or Python, or R. Based on observations and given priors, in the same loop construct two Gibbs samplers, one for (µ1, τ1) and
the other for (µ2, τ2).
2Gerhardt, H. C. (1994). Reproductive character displacement of female mate choice in the grey treefrog,
Hyla chrysoscelis. Anim. Behav., 47, 959–969.
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Form a sequence of differences µ1,j − µ2,j , j = 1, . . . , 11000, and after rejecting the initial
1000 differences, from the remaining simulations estimate 95% credible set for µ1 − µ2.
Does this set contain zero? What can you say about the hypothesis H0 : µ1 = µ2
based on this credible set? Based on this analysis elaborate whether the length of call is a
discriminatory characteristic?
Hint: When no raw data are given, that is, when data are summarized via sample size,
sample mean, and sample standard deviation, the following identity may be useful:
Xn
i=1
(yi − µ)
2 = (n − 1)s
2 + n(y − µ)
2
,
where n, y, and s are sample size, sample mean, and sample standard deviation, respectively.
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