Distribution of $Z^2 \cdot I(Z > 0)$ where $Z \sim \text{N}(0,1)$

Question

When using the Likelihood Ratio test for testing particular hypotheses and attempting to obtain an size-$\alpha$ test, I run into the expression $$ \mathbb{P}\left( Z^2 \cdot I(Z > 0) > c \right) = \alpha $$ where $I(\cdot)$ is the typical indicator function. My goal is to find an expression for $c$ so that the above statement is true. I want to check if my reasoning is correct for finding the distribution of this $Z^2 \cdot I(Z > 0)$ random variable.

Let $Y = Z^2 \cdot I(Z > 0) = Z^2 \vert (Z > 0)$. Then

\begin{align*} \mathbb{P}(Y \leq y) &= \mathbb{P}(Z^2 \leq y \vert Z > 0) \\[8pt] &= \frac{\mathbb{P}(Z^2 \leq y \quad\text{ AND }\quad Z > 0)}{\mathbb{P}(Z > 0)} \\[8pt] &= 2 \mathbb{P}( -y \leq Z \leq y \quad\text{ AND }\quad Z > 0) \\[8pt] &= 2 \mathbb{P}(0 < Z \leq y) \\[8pt] &= 2 \Phi(y) - 1 \end{align*} where $\Phi(x)$ is the standard normal CDF. From this expression, I think I then may be able to find the desired $c$ through some algebra. Is this reasoning correct for finding the distribution of $Z^2 \vert (Z > 0)$?

Answer 1

For the problem "finding $c$ such that $P(Z^2I(Z > 0) > c) = \alpha$", it is entirely unnecessary to introduce any conditional notation or argument (please see my comment above too). It is just a simple application of the law of total probability.

To begin with, $c$ must be non-negative, for the random variable $Z^2I(Z > 0)$ is non-negative hence if $c$ is negative, then $\alpha = 1$, which is impractical (assuming $\alpha$ is the size of some hypothesis testing problem). That said, by the law of total probability: \begin{align*} & P(Z^2I(Z > 0) > c) \\ ={}& P(Z^2I(Z > 0) > c, Z > 0) + P(Z^2I(Z > 0) > c, Z \leq 0) \\ ={}& P(Z^2 > c, Z > 0) + P(0 > c, Z \leq 0) \\ ={}& P(Z > \sqrt{c}) = 1 - \Phi(\sqrt{c}). \end{align*} The last equality is due to $Z \sim N(0, 1)$. Solving $1 - \Phi(\sqrt{c}) = \alpha$ then gives $$c = (\Phi^{-1}(1 - \alpha))^2,$$ where $\Phi^{-1}(p)$ denotes the $p$-quantile of $\Phi$.

Answer 2

You can argue that $Z^2 \mid (Z > 0) \sim Z^2 \mid (Z \leq 0)$ and hence also $Z^2 \mid (Z > 0) \sim Z^2$. So the distribution is chi-squared with one degree of freedom.

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Edit: The question has been changed from being about $Z^2 \mid (Z > 0)$ to $Z^2 \cdot I(Z > 0)$. In this case you get a mixture of a chi-squared distribution with one degree of freedom, and a point mass at zero.

Answer 3

Since the distribution of $Z$ is symmetric around zero, you have:

$$\begin{align} \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z > 0) &= \frac{\mathbb{P}(0 \leqslant Z^2 \leqslant t, Z > 0)}{\mathbb{P}(Z > 0)} \\[6pt] &= \frac{\mathbb{P}(0 \leqslant |Z| \leqslant \sqrt{t}, Z > 0)}{\mathbb{P}(Z > 0)} \\[6pt] &= \frac{\mathbb{P}(0 \leqslant |Z| \leqslant \sqrt{t}, Z < 0)}{\mathbb{P}(Z < 0)} \\[6pt] &= \frac{\mathbb{P}(0 \leqslant Z^2 \leqslant t, Z < 0)}{\mathbb{P}(Z < 0)} \\[12pt] &= \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z < 0) \\[6pt] \end{align}$$

You therefore have:

$$\begin{align} \mathbb{P}(0 \leqslant Z^2 \leqslant t) &= \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z > 0) \cdot \mathbb{P}(Z > 0) +\mathbb{P}(0 \leqslant Z^2 \leqslant t| Z < 0) \cdot \mathbb{P}(Z < 0) + \mathbb{P}(Z = 0) \\[12pt] &= \frac{1}{2} \bigg[ \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z > 0) + \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z < 0) \bigg] + \mathbb{P}(Z = 0) \\[6pt] &= \frac{1}{2} \times 2 \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z > 0) + \mathbb{P}(Z = 0) \\[12pt] &= \mathbb{P}(0 \leqslant Z^2 \leqslant t| Z > 0) + \mathbb{P}(Z = 0). \\[6pt] \end{align}$$

This general rule holds for any symmetric distribution around zero for $Z$. If the distribution is continuous then we have $\mathbb{P}(Z = 0) = 0$ which then means that $Z^2 \sim Z^2|(Z>0)$. This equivalence is enough to establish that $Z^2|(Z>0) \sim \text{ChiSq}(1)$ in the present case.

Answer 4

For completeness, let me note that this mixture distribution is known in (particle) physics as the half chi-squared distribution and denoted by $\tfrac12 \chi^2_1$.

See the top of p15 in 1007.1727 for further discussion, including similar equations to those in the original question.

It is relevant in the context of physics as we seek non-negative signals. This leads to LLR test-statistics that asymptotically follow a half chi-squared distribution rather than a chi-squared distribution. This result is Chernoff's theorem, which is a modification of Wilks' theorem.