# Help with understanding Turing's predict function

**URL:** <https://discourse.julialang.org/t/help-with-understanding-turings-predict-function/102021>\
**Category:** Probabilistic Programming\
**Tags:** question, turing\
**Created:** [July 24, 2023, 1:52pm UTC](https://discourse.julialang.org/t/help-with-understanding-turings-predict-function/102021 "2023-07-24T13:52:27Z")\
**Posts on this page:** 4\
**Page:** 1

<div class="post-metadata">

**Author:** ![JanG](https://avatars.discourse-cdn.com/v4/letter/j/ecb155/32.png) [@JanG](https://discourse.julialang.org/u/JanG)\
**Post date:** [July 24, 2023, 1:52pm UTC](https://discourse.julialang.org/t/help-with-understanding-turings-predict-function/102021/1 "2023-07-24T13:52:28Z")

</div>

Hi everybody,

I played around a bit with the `predict()` function in Turing and encountered some behaviour I don’t quite understand. When I use it to produce new output based on a chain, it only seems to take some of the variables in the original chain into account but not others.

Here is some example codes that reproduces the problem. In the code I simulate some data for a simple linear regression, and then fit that data using Turing. I then use the `predict` function to generate new data based on the fitted chain, and fit the predicted data once more using the same model.

```julia
using Turing, StatsPlots

@model function linear_reg(x, y)
    β ~ Normal(0, 1)
    σ ~ Exponential(0.1)

    for i ∈ eachindex(y)
        y[i] ~ Normal(β * x[i], σ)
    end
end;

pars = Array{Float64}(undef, 20, 2, 2)

for i = 1:20
    # simulate some data and fit a model to it
    f(x) = (i-20)/10 * x + i/100 * randn();
    Δ = 0.1; xs_train = 0:Δ:100; ys_train = f.(xs_train);
    m_train = linear_reg(xs_train, ys_train);
    chain_lin_reg = sample(m_train, NUTS(), 1000);

    # make predictions based on the fitted model
    m_test = linear_reg(xs_train, Vector{Union{Missing, Float64}}(undef, length(ys_train)));
    predictions = predict(m_test, chain_lin_reg)
    ys_predictions = vec(mean(Array(group(predictions, :y)); dims = 1))
    
    # fit a model to the predictions
    m_recover = linear_reg(xs_train, ys_predictions);
    chain_lin_reg_recover = sample(m_recover, NUTS(), 1000);

    # store the parameters
    pars[i, 1, :] = mean(Array(chain_lin_reg), dims = 1)
    pars[i, 2, :] = mean(Array(chain_lin_reg_recover), dims = 1)
end

p1 = scatter(pars[:,1,1], pars[:,2,1], legend=false, aspect_ratio=:equal);
title!("β")
p2 = scatter(pars[:,1,2], pars[:,2,2], legend=false, aspect_ratio=:equal);
title!("σ")
plot(p1, p2, layout = (1,2), legend=false)
xlabel!("original")
ylabel!("recovered")

```

![prediciton_troubles](https://global.discourse-cdn.com/julialang/original/3X/f/1/f19e214b9746aef7935c348597b4258052193dc8.png)

As can be seen from the plot this code produces, the slope parameter of the regression model seems to be used in generating data when using `predict` but the noise parameter is not used for some reason. Is there something I’ve done wrong or I’m missing here, and if not, is this the intended behaviour of `predict`? Thanks for your help!

---

<div class="post-metadata">

**Author:** ![math4mad](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/math4mad/32/49486_2.png) [@math4mad](https://discourse.julialang.org/u/math4mad)\
**Post date:** [July 30, 2023, 4:43pm UTC](https://discourse.julialang.org/t/help-with-understanding-turings-predict-function/102021/2 "2023-07-30T16:43:52Z")

</div>

you want like this ?

 ![posterior params](https://global.discourse-cdn.com/julialang/original/3X/8/1/81a770ce63013661e5845068f4151d9f8da8e81e.png)

---

<div class="post-metadata">

**Author:** ![JanG](https://avatars.discourse-cdn.com/v4/letter/j/ecb155/32.png) [@JanG](https://discourse.julialang.org/u/JanG)\
**Post date:** [July 30, 2023, 5:11pm UTC](https://discourse.julialang.org/t/help-with-understanding-turings-predict-function/102021/3 "2023-07-30T17:11:23Z")

</div>

I would’ve expected sigma to also be applied to the output of `predict`. Here I’m doing it manually to show what I expected the result to look like:

```julia
using Turing, StatsPlots

@model function linear_reg(x, y)
    β ~ Normal(0, 1)
    σ ~ Exponential(0.1)

    for i ∈ eachindex(y)
        y[i] ~ Normal(β * x[i], σ)
    end
end;

pars = Array{Float64}(undef, 20, 2, 2)

for i = 1:10
    # simulate some data and fit a model to it
    f(x) = (i-5)/10 * x + i/100 * randn();
    Δ = 0.1; xs_train = 0:Δ:100; ys_train = f.(xs_train);
    m_train = linear_reg(xs_train, ys_train);
    chain_lin_reg = sample(m_train, NUTS(), 1000);

    # make predictions based on the fitted model
    m_test = linear_reg(xs_train, Vector{Union{Missing, Float64}}(undef, length(ys_train)));
    predictions = predict(m_test, chain_lin_reg)
    ys_predictions = vec(mean(Array(group(predictions, :y)); dims = 1))

    # add noise to the predictions
    σ = mean(Array(group(chain_lin_reg, :σ)); dims = 1)
    ys_predictions = rand.(Normal.(ys_predictions, σ))

    # fit a model to the predictions
    m_recover = linear_reg(xs_train, ys_predictions);
    chain_lin_reg_recover = sample(m_recover, NUTS(), 1000);

    # store the parameters
    pars[i, 1, :] = mean(Array(chain_lin_reg), dims = 1)
    pars[i, 2, :] = mean(Array(chain_lin_reg_recover), dims = 1)
end

p1 = scatter(pars[:,1,1], pars[:,2,1], legend=false, aspect_ratio=:equal);
title!("β")
p2 = scatter(pars[:,1,2], pars[:,2,2], legend=false, aspect_ratio=:equal);
title!("σ")
fig = plot(p1, p2, layout = (1,2), legend=false)
xlabel!("original")
ylabel!("recovered")
savefig(fig,"prediciton_troubles.png")

```

![prediciton_troubles](https://global.discourse-cdn.com/julialang/original/3X/3/d/3dccc85aeae1fcf226a269cc0eb1e3800eb101f5.png)

---

<div class="post-metadata">

**Author:** ![torfjelde](https://sea2.discourse-cdn.com/julialang/user_avatar/discourse.julialang.org/torfjelde/32/206542_2.png) [@torfjelde](https://discourse.julialang.org/u/torfjelde)\
**Post date:** [September 2, 2023, 9:42am UTC](https://discourse.julialang.org/t/help-with-understanding-turings-predict-function/102021/4 "2023-09-02T09:42:28Z")

</div>

The issue here is that you’re using the _mean_ prediction in the first example:)

The following is your offending line:

```julia
    ys_predictions = vec(mean(Array(group(predictions, :y)); dims = 1))

```

If you actually inspect `predictions`, i.e. the chain from the line before, you can see that the predictions do indeed have the correct std. Here is the result for `i = 10`:

```julia
...
Summary Statistics
  parameters mean std mcse ess_bulk ess_tail rhat ess_per_sec 
      Symbol Float64 Float64 Float64 Float64 Float64 Float64 Missing 

        y[1] -0.0013 0.0966 0.0030 1046.1171 868.7508 1.0004 missing
        y[2] -0.0980 0.1018 0.0034 907.8375 845.5306 1.0008 missing
        y[3] -0.1987 0.0985 0.0033 903.8513 784.7725 1.0002 missing
        y[4] -0.2984 0.0970 0.0030 995.8047 982.7992 0.9993 missing
        y[5] -0.4002 0.0989 0.0031 1046.4576 942.0177 0.9999 missing
        y[6] -0.5024 0.0980 0.0032 966.1871 767.5986 1.0021 missing
        y[7] -0.6039 0.0972 0.0031 1014.4943 1024.3146 0.9999 missing
        y[8] -0.7007 0.1003 0.0031 1046.7164 942.2477 0.9990 missing
...

```

As you can see, the std. is correct.

But then in the above line you just average this out:) And so you’re really just trying to fit the mean regression, and hence you get ~0 variance in the posterior.