GATE 2026 DA Q.28: solution and answer

Official answer: A, CSource: Official answer key, GATE 2026 DA (IIT Guwahati)

1. A $t_1$ distribution is the standard Cauchy distribution: $h(y)=\frac{1}{\pi (1+y^2)}$ and $H(y)=\frac{1}{2}+\frac{\arctan y}{\pi }$. The normal density is $g(z)=\frac{1}{\sqrt{2\pi }}{e}^{-z^2/2}$.

2. (A) Both densities are symmetric about $0$, so each CDF equals $\frac{1}{2}$ at $0$: $G(0)=H(0)=\frac{1}{2}$. True.

3. (D) $g(0)=\frac{1}{\sqrt{2\pi }}\approx 0.399$ while $h(0)=\frac{1}{\pi }\approx 0.318$. They differ, so (D) is false.

4. Locate $c$. The condition $g(c)=h(c)$ is equivalent to $(1+c^2)e^{-c^2/2}=\sqrt{2\pi }/\pi \approx 0.798$. The left side equals $1$ at $c=0$, rises to about $1.21$ at $c=1$, then decreases to $0$, so it meets $0.798$ exactly once for $c>0$. At $c=1$ we have $g=0.242>h=0.159$ and at $c=2$ we have $g=0.054<h=0.064$, so $1<c<2$. Solving numerically gives $c\approx 1.85$, where $g\approx 0.0721$ and $h\approx 0.0720$. For $z>c$ the Cauchy density stays larger, because its tail decays polynomially while the normal tail decays like $e^{-z^2/2}$.

5. (B) $G(c)\approx 0.968$, while $H(c)=\frac{1}{2}+\frac{\arctan 1.85}{\pi }\approx 0.5+0.343=0.843$. So $G(c)>H(c)$ and (B) is false. Equivalently, the Cauchy has more mass in the tail $(c,\infty )$, so less mass below $c$.

6. (C) By symmetry $G(-c)=1-G(c)\approx 0.032$ and $H(-c)=1-H(c)\approx 0.157$. So $G(-c)<H(-c)$ and (C) is true. This is the same tail comparison, seen on the left.

7. Correct statements: (A) and (C).