Exercise 11.15 - Posterior mean and variance of a truncated Gaussian

Answers

Denote $A = \frac{c_{i} - μ_{i}}{σ}$ , for the conditional mean, by linearity:

𝔼 [z_{i} | z_{i} \geq c_{i}] = μ_{i} + σ \cdot 𝔼 [𝜖_{i} | 𝜖_{i} \geq A] .

And we have:

𝔼 [𝜖_{i} | 𝜖_{i} \geq A] = \frac{1}{p (𝜖_{i} \geq A)} \cdot \int_{A}^{+ \infty} 𝜖_{i} \cdot 𝒩 (𝜖_{i} | 0, 1) d 𝜖_{i} = \frac{ϕ (A)}{1 - Φ (A)} = H (A),

where $H$ is defined by (11.139). (Recall the definition of the conditional expectation!) Therefore we have:

𝔼 [z_{i} | z_{i} \geq c_{i}] = μ_{i} + σ \cdot H (A) .

Now we proceed to calculate the expectation for the squared term:

𝔼 [z_{i}^{2} | z_{i} \geq c_{i}] = μ_{i}^{2} + 2 μ_{i} 𝜎𝔼 [𝜖_{i} | 𝜖_{i} \geq A] + σ^{2} 𝔼 [𝜖_{i}^{2} | 𝜖_{i} \geq A] .

To evaluate $𝔼 [𝜖_{i}^{2} | 𝜖_{i} \geq A]$ , we make use of the hint of this exercise:

\frac{d}{d w} (w \cdot 𝒩 (w | 0, 1)) = 𝒩 (w | 0, 1) - w^{2} \cdot 𝒩 (w | 0, 1) .

Hence we can solve for the following integral by parts:

\int_{b}^{c} w^{2} \cdot 𝒩 (w | 0, 1) d w = Φ (c) - Φ (b) - c \cdot 𝒩 (c | 0, 1) + b \cdot 𝒩 (b | 0, 1) .

Thence:

𝔼 [𝜖_{i}^{2} | 𝜖_{i} \geq A] = \frac{1}{p (𝜖_{i} \geq A)} \cdot \int_{A}^{+ \infty} w^{2} \cdot 𝒩 (w | 0, 1) d w = \frac{1 - Φ (A) + A \cdot ϕ (A)}{1 - Φ (A)} .

Plug it into the previous formula:

\begin{aligned} 𝔼 [z_{i}^{2} | z_{i} \geq c_{i}] & = μ_{i}^{2} + 2 μ_{i} σ \cdot H (A) + σ^{2} \frac{1 - Φ (A) + 𝐴𝜙 (A)}{1 - Φ (A)} \\ = μ_{i}^{2} + σ^{2} + H (A) (σ c_{i} + σ μ_{i}) . \end{aligned}

solour_lfq

2021-03-24 13:42