The attached file contains experimental data showing a spectral line observed in a laboratory. Your task is to perform a Bayesian analysis on a grid to adjust the amplitude A of the spectral line and the background flux B. Assume that the spectral line has a Gaussian shape. For physical reasons, the value of A is constrained to be between 0 and 1000 a priori. The value of B is constrained to be between 0 and 3000 a priori. Use a uniform “prior” on A and B on a linear scale.

1 Exercise 2

  • Test the function provided to you for the physics model of the spectral line. What is the vector of return values for A=200, B=1860? Copy the values into your document.
Code 
using Distributions
using CairoMakie

# Define the model function
function model(A, B, x; xpos=10)
    d = Normal(xpos, 1.0)
    return B .+ A .* pdf.(d, x)
end

# Define the data
data = [1837.00283015, 1811.76260793, 1877.0602172, 1923.04345098,
    1810.09024628, 1898.42037995, 1784.35363285, 1866.72588538,
    1826.22446702, 1896.58934977, 2043.89952533, 1946.20171261,
    1842.25394544, 1803.83175398, 1803.74949885, 1844.61491832,
    1828.0670997, 1816.13580618, 1885.88536203, 1846.00959492]

# Set up the x values
l = length(data)
x = 0:(l-1)

# Test the model function with A=200, B=1860
A = 200
B = 1860
model_values = model(A, B, x)
println("Model values for A=200, B=1860:")
println(model_values)

# Plot the data and model
f = Figure()
ax = Axis(f[1, 1], xlabel="Pixel", ylabel="Photons Collected")
stairs!(ax, x, data, label="Data")
stairs!(ax, x, model_values, label="Model (A=$(A), B=$(B))")
f
Model values for A=200, B=1860:
[1860.0, 1860.0, 1860.000000000001, 1860.000000001827, 1860.0000012151766, 1860.000297343903, 1860.026766045153, 1860.8863696823876, 1870.7981933026376, 1908.3941449038286, 1939.7884560802866, 1908.3941449038286, 1870.7981933026376, 1860.8863696823876, 1860.026766045153, 1860.000297343903, 1860.0000012151766, 1860.000000001827, 1860.000000000001, 1860.0]

  • Code a function “prior” returning prior pdf p(A,B|I) for a given set of parameters (A, B). What is the return value for A=200, B=1860? Ideally, write the function in a vectorized form that can return the likelihood values with A and B values given on 2D grids.
Code 
# Define the prior function
function prior(A, B)
    if 0 <= A <= 1000 && 0 <= B <= 3000
        return 1.0 / (1000 * 3000)  # Normalized uniform prior
    else
        return 0.0
    end
end

prior_value = prior(A, B)
println("Prior value for A=200, B=1860: ", prior_value)
Prior value for A=200, B=1860: 3.3333333333333335e-7
  • Code a function “like” returning the likelihood function p(data|A,B,I) for a given set of parameters. What is the return value for A=200, B=1860? Ideally, write the function in a vectorized form that can return the likelihood values with A and B values given on 2D grids.
Code 
# Define the likelihood function
function like(A, B, data, x; xpos=10, sigma=30.0)
    # Calculate the model prediction
    m = model(A, B, x, xpos=xpos)
    # Calculate the likelihood assuming Gaussian errors
    probs = @. pdf(Normal(m, sigma), data)
    return prod(probs)
end

# Test the likelihood function with A=200, B=1860
likelihood_value = like(A, B, data, x)
println("Likelihood value for A=200, B=1860: ", likelihood_value)
Likelihood value for A=200, B=1860: 1.6409575671741215e-48
  • Code a function “like_times_prior” returning the products of the values of the likelihood and prior for a given set of parameters. That is the numerator of Bayes’ Law. What is the return value for A=200, B=1860?
Code 
# Define the function for the numerator of Bayes' Law
function like_times_prior(A, B, data, x; xpos=10, sigma=30.0)
    return like(A, B, data, x, xpos=xpos, sigma=sigma) * prior(A, B)
end

# Test the function with A=200, B=1860
posterior_numerator = like_times_prior(A, B, data, x)
println("Posterior numerator for A=200, B=1860: ", posterior_numerator)
Posterior numerator for A=200, B=1860: 5.469858557247072e-55

2 Exercise 3:

  1. Add to the code to calculate the joint posterior probability density p(A,B|data,I) of A and B on a 300x300 grid spanning A and B. Note: Use np.meshgrid to generate a 2D grid for the values of A and B.
Code 
# Create a grid of A and B values
f = Figure()
A_values = range(0, 900, length=200) |> collect
B_values = range(1780, 1940, length=300) |> collect

# Calculate the posterior (unnormalized) for each grid point
posterior_grid = zeros(length(A_values), length(B_values))
for (i, A) in enumerate(A_values)
    for (j, B) in enumerate(B_values)
        posterior_grid[i, j] = like_times_prior(A, B, data, x)
    end
end

posterior_grid = posterior_grid ./ sum(posterior_grid)

# Find the maximum posterior and its location
max_posterior = maximum(posterior_grid)
max_idx = argmax(posterior_grid)
max_A = A_values[max_idx[1]]
max_B = B_values[max_idx[2]]

println("Maximum posterior value: ", max_posterior)
println("Maximum posterior at A = ", max_A, ", B = ", max_B)

# Plot the posterior
ax = Axis(f[1, 1], xlabel="Amplitude (A)", ylabel="Background (B)", title="Posterior Probability")
hm = heatmap!(ax, A_values, B_values, (posterior_grid))
Colorbar(f[1, 1][1, 2], hm)
f
Maximum posterior value: 0.0009218105353821632
Maximum posterior at A = 434.1708542713568, B = 1837.7926421404682

  1. Make a graph of the posterior distribution in 2D using ax.pcolormesh(). Label the axes. Use a 300x300 grid over the entire parameter space. Label the axes. Include a color bar with the label.
  2. Draw the same graph as in a) but use a 300x300 grid spanning only the non-zero probability region. Label the axes. Include a color bar with the label.
  3. Show the point of maximum probability with a star symbol (marker=“*”) on the posterior of 2c). Include a color bar with the label.
  4. Make a new graph identical to the one in Step 1 and show the best fit by superimposing the best fit model on the data. Write down the maximum probability values of A and B.