The problem asks for the value of C such that the power of the suggested test is 0.5.

The test statistic is $T(X) = \frac{X}{\sqrt{1+X^2}}$. The hypothesis is $H_0: \theta = -1$ versus $H_1: \theta = 0$. The test rejects $H_0$ if $T(X) > C$. The power of the test is defined as the probability of rejecting $H_0$ when $H_1$ is true.

Power = $P(\text{Reject } H_0 | H_1 \text{ is true}) = P\left(\frac{X}{\sqrt{1+X^2}} > C | \theta = 0\right)$.

We are given that the power should be 0.5.

Distribution of the Test Statistic under H₁

Under $H_1$, we have $\theta = 0$. The random variable $X$ follows a Cauchy distribution with location parameter 0 and scale 1, i.e., $X \sim \text{Cauchy}(0, 1)$. The probability density function (PDF) is $f_0(x) = \frac{1}{\pi (1+x^2)}$.

Let $Y = \frac{X}{\sqrt{1+X^2}}$. We need to find the distribution of $Y$ when $X \sim \text{Cauchy}(0, 1)$. A convenient transformation is $X = \tan(Z)$, where $Z$ follows a uniform distribution on $(-\frac{\pi}{2}, \frac{\pi}{2})$, i.e., $Z \sim U(-\frac{\pi}{2}, \frac{\pi}{2})$.

Under this transformation, the test statistic becomes:

$Y = \frac{\tan(Z)}{\sqrt{1+\tan^2(Z)}} = \frac{\tan(Z)}{\sec(Z)} = \sin(Z)$.

Since $Z \sim U(-\frac{\pi}{2}, \frac{\pi}{2})$, its PDF is $f_Z(z) = \frac{1}{\pi}$ for $z \in (-\frac{\pi}{2}, \frac{\pi}{2})$. The range of $Y = \sin(Z)$ is $(-1, 1)$.

The PDF of $Y$ can be found using the transformation $Y = \sin(Z)$. The CDF of $Y$ is $F_Y(y) = P(Y \le y) = P(\sin(Z) \le y)$. For $y \in (-1, 1)$, $P(\sin(Z) \le y) = P(Z \le \arcsin(y)) = \int_{-\pi/2}^{\arcsin(y)} \frac{1}{\pi} dz = \frac{1}{\pi} (\arcsin(y) + \frac{\pi}{2})$.

The PDF of $Y$ is the derivative of the CDF:

$f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{1}{\pi} \frac{1}{\sqrt{1-y^2}}$, for $y \in (-1, 1)$.

Calculating the Value of C

We need to find $C$ such that $P(Y > C) = 0.5$. Using the PDF of $Y$:

$ P(Y > C) = \int_C^1 f_Y(y) dy = \int_C^1 \frac{1}{\pi \sqrt{1-y^2}} dy $

Set this equal to 0.5:

$ \int_C^1 \frac{1}{\pi \sqrt{1-y^2}} dy = 0.5 $

Evaluate the integral:

$ \frac{1}{\pi} [\arcsin(y)]_C^1 = 0.5 $

$ \frac{1}{\pi} (\arcsin(1) - \arcsin(C)) = 0.5 $

$ \frac{1}{\pi} \left(\frac{\pi}{2} - \arcsin(C)\right) = 0.5 $

Multiply both sides by $\pi$:

$ \frac{\pi}{2} - \arcsin(C) = \frac{\pi}{2} $

Solve for $\arcsin(C)$:

$ \arcsin(C) = 0 $

Therefore, $C = \sin(0) = 0$.

Alternative Method (Symmetry)

The PDF $f_Y(y) = \frac{1}{\pi \sqrt{1-y^2}}$ is symmetric about $y=0$. Since the total probability $\int_{-1}^1 f_Y(y) dy = 1$, the median of the distribution is 0.

The condition for the power to be 0.5 is $P(Y > C) = 0.5$. Since 0 is the median, $P(Y > 0) = 0.5$. Therefore, $C=0$.