Question:
Suppose a random variable \(Z\) follows \(Normal(\mu=0, \sigma^2=1)\) distribution with probability density function \(g(z)\) and cumulative distribution function \(G(z)\). Another random variable \(Y\) follows \(t_1\) distribution with probability density function \(h(y)\) and cumulative distribution function \(H(y)\). Let \(c\) be the positive real number for which \(g(c) = h(c)\).
Which of the following statements is/are correct?
Suppose a random variable \(Z\) follows \(Normal(\mu=0, \sigma^2=1)\) distribution with probability density function \(g(z)\) and cumulative distribution function \(G(z)\). Another random variable \(Y\) follows \(t_1\) distribution with probability density function \(h(y)\) and cumulative distribution function \(H(y)\). Let \(c\) be the positive real number for which \(g(c) = h(c)\).
Which of the following statements is/are correct?
Which of the following statements is/are correct?
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Remember both densities are symmetric about 0, so their CDFs must equal 0.5 there regardless of the shape of the tails.
Updated On: Jul 22, 2026
- \(G(0) = H(0)\)
- \(G(c) < H(c)\)
- \(G(-c) < H(-c)\)
- \(g(0) = h(0)\)
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The Correct Option is A, C
Solution and Explanation
Step 1: Note the shape of both densities.
\(Z\) follows a standard normal distribution and \(Y\) follows a \(t_1\) distribution (the same as a standard Cauchy distribution). Both densities \(g(z)\) and \(h(y)\) are symmetric about 0, bell-shaped, and strictly decreasing as you move away from 0 in either direction.
Step 2: Compare the peak heights at 0.
\[ g(0) = \frac{1}{\sqrt{2\pi}} \approx 0.399, \qquad h(0) = \frac{1}{\pi(1+0^2)} = \frac{1}{\pi} \approx 0.318 \]
So \(g(0) > h(0)\), meaning statement (D), \(g(0)=h(0)\), is FALSE. The normal curve is taller at the centre than the Cauchy curve.
Step 3: Use symmetry to compare the CDFs at 0.
Both densities are symmetric about 0, so exactly half the probability mass lies on each side of 0 for both distributions. This gives \(G(0) = 0.5\) and \(H(0) = 0.5\), so \(G(0) = H(0)\). Statement (A) is TRUE.
Step 4: Understand where c comes from.
The normal density falls off very fast (like \(e^{-z^2/2}\)) while the Cauchy density falls off slowly (like \(1/y^2\)), so the Cauchy has much heavier tails. Since \(g(0) > h(0)\) but the Cauchy eventually overtakes the normal far from 0, the two curves must cross at some positive point, and this crossing point is \(c\). For \(0 < y < c\), \(g(y) > h(y)\) (normal is higher, closer to the centre); for \(y > c\), \(h(y) > g(y)\) (Cauchy's heavy tail is higher).
Step 5: Check option (B), G(c) < H(c).
\[ G(c) - G(0) = \int_0^c g(y)\,dy, \qquad H(c) - H(0) = \int_0^c h(y)\,dy \]
Since \(g(y) > h(y)\) throughout \((0,c)\), the left integral is larger than the right one. Because \(G(0)=H(0)=0.5\), this means \(G(c) > H(c)\), not less than. So statement (B) is FALSE.
Step 6: Check option (C), G(-c) < H(-c).
By symmetry of both distributions about 0, \(G(-c) = 1 - G(c)\) and \(H(-c) = 1 - H(c)\). Since \(G(c) > H(c)\) from Step 5, subtracting both sides from 1 flips the inequality: \(1 - G(c) < 1 - H(c)\), which is \(G(-c) < H(-c)\). So statement (C) is TRUE.
Final Answer:
Statements (A) and (C) are correct. \[ \boxed{\text{(A) and (C)}} \]
\(Z\) follows a standard normal distribution and \(Y\) follows a \(t_1\) distribution (the same as a standard Cauchy distribution). Both densities \(g(z)\) and \(h(y)\) are symmetric about 0, bell-shaped, and strictly decreasing as you move away from 0 in either direction.
Step 2: Compare the peak heights at 0.
\[ g(0) = \frac{1}{\sqrt{2\pi}} \approx 0.399, \qquad h(0) = \frac{1}{\pi(1+0^2)} = \frac{1}{\pi} \approx 0.318 \]
So \(g(0) > h(0)\), meaning statement (D), \(g(0)=h(0)\), is FALSE. The normal curve is taller at the centre than the Cauchy curve.
Step 3: Use symmetry to compare the CDFs at 0.
Both densities are symmetric about 0, so exactly half the probability mass lies on each side of 0 for both distributions. This gives \(G(0) = 0.5\) and \(H(0) = 0.5\), so \(G(0) = H(0)\). Statement (A) is TRUE.
Step 4: Understand where c comes from.
The normal density falls off very fast (like \(e^{-z^2/2}\)) while the Cauchy density falls off slowly (like \(1/y^2\)), so the Cauchy has much heavier tails. Since \(g(0) > h(0)\) but the Cauchy eventually overtakes the normal far from 0, the two curves must cross at some positive point, and this crossing point is \(c\). For \(0 < y < c\), \(g(y) > h(y)\) (normal is higher, closer to the centre); for \(y > c\), \(h(y) > g(y)\) (Cauchy's heavy tail is higher).
Step 5: Check option (B), G(c) < H(c).
\[ G(c) - G(0) = \int_0^c g(y)\,dy, \qquad H(c) - H(0) = \int_0^c h(y)\,dy \]
Since \(g(y) > h(y)\) throughout \((0,c)\), the left integral is larger than the right one. Because \(G(0)=H(0)=0.5\), this means \(G(c) > H(c)\), not less than. So statement (B) is FALSE.
Step 6: Check option (C), G(-c) < H(-c).
By symmetry of both distributions about 0, \(G(-c) = 1 - G(c)\) and \(H(-c) = 1 - H(c)\). Since \(G(c) > H(c)\) from Step 5, subtracting both sides from 1 flips the inequality: \(1 - G(c) < 1 - H(c)\), which is \(G(-c) < H(-c)\). So statement (C) is TRUE.
Final Answer:
Statements (A) and (C) are correct. \[ \boxed{\text{(A) and (C)}} \]
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