The very last part of theorem 5.4(pg.356) of TPE by Lehmann and Casella

I am struggling with the very last part of theorem 5.4(pg.356) of TPE by Lehmann and Casella.

It is the theorem that proves the inadmissibility of the James Stein Estimator dominated by the positive part James Stein Estimator.

What I am struggling with is the last part of the proof of the theorem.

The followings are the TWO PARTS that I am struggling with.

1.

I can't derive the last equality from the bottom third line of the proof.

I've carefully read the explanation(by S.Catterall Reinstate Monica) but I am still confused whether it is possible to derive the conditional expectation of continuous random variable by treating it as discrete.

2.

Also I can't understand the bottom two lines of the proof.

I've found that it is an increasing function of the absolute value of |$$\theta_1 y$$|, but I can't understand how it completes the proof.

Thank you.

Part 1. The equality towards the end of the proof requires computing a conditional expectation $$E_\theta (X_1 | X_1^2=y^2)$$ where, earlier in the text, we are told that $$X_1$$ is normally distributed with mean $$\theta_1$$ and variance $$1$$ ($$\theta_1$$ being the first component of the vector $$\theta$$). The condition $$X_1^2=y^2$$ implies that either $$X_1=y$$ or $$X_1=-y$$. The conditional probability that $$X_1$$ takes value $$y$$ is equal to $$\frac{f(y)}{f(y)+f(-y)}$$ where $$f$$ is the (normal) pdf of $$X_1$$. So $$E_\theta (X_1 | X_1^2=y^2)=y\frac{f(y)}{f(y)+f(-y)}+(-y)\frac{f(-y)}{f(y)+f(-y)}$$. Substitute the formula for $$f$$ into this and you get the expression stated in the printed proof, namely $$E_\theta (X_1 | X_1^2=y^2)=y\frac{e^{\theta_1 y}-e^{-\theta_1 y}}{e^{\theta_1 y}+e^{-\theta_1 y}}$$

Note: this is an intuitive derivation of the conditional expectation. For a more rigorous approach see this answer. The answer uses a lemma to obtain a certain function $$g$$; this lemma is proved on pages 206-207 of the book Probability with Martingales.

Part 2. The computed conditional expectation is found to be an increasing function of $$|\theta_1 y|$$ which is zero when $$\theta_1 y=0$$. This means that it must be non-negative. Finally, $$E_\theta (\theta_1 X_1 | \hat{B})=E(E(\theta_1 X_1|\hat{B}, X_2,\cdots,X_r)|\hat{B})$$ by the tower property for conditional expectations (pages 88-89 in the above book) so $$E_\theta (\theta_1 X_1 | \hat{B})$$ is also non-negative, which is what they were trying to prove.

• I’ve got one more extra question for your explanation. I’m confusing how it is possible to regard the probability density function like probability mass function since I can’t get the rigorous reason why the conditional probability that X1 takes value y is equal to the formula that you’ve mentioned. I’ve got the intuition, but would you mind if I ask you to give more clearer explanation on that point? Thank you. – Kyungmin Kim Nov 20 '20 at 0:57
• Thank you so much. I really appreciate your explanation with care and clarity. Thanks again once more:) – Kyungmin Kim Nov 20 '20 at 14:12