Question:
If \(X_1\) and \(X_2\) are two independent random variables having identical N(0, 1) distribution, then \(P(X_1^2 + X_2^2 \le 2)\) equals:
If \(X_1\) and \(X_2\) are two independent random variables having identical N(0, 1) distribution, then \(P(X_1^2 + X_2^2 \le 2)\) equals:
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\(X_1^2+X_2^2\) follows a chi-square distribution with 2 df, whose CDF is \(1-e^{-x/2}\).
Updated On: Jul 4, 2026
- \(e^{-1}\)
- \(e^{-2}\)
- \(1-e^{-1}\)
- \(1-e^{-2}\)
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The Correct Option is C
Solution and Explanation
Step 1: If \(X_1, X_2\) are independent standard normal variables, then \(X_1^2\) and \(X_2^2\) are each chi-square with 1 degree of freedom, and their sum \(X_1^2 + X_2^2\) follows a chi-square distribution with 2 degrees of freedom:\[ X_1^2 + X_2^2 \sim \chi^2_2 \]
Step 2: A chi-square distribution with 2 degrees of freedom is the same as an Exponential distribution with mean 2 (rate \(\lambda = 1/2\)). Its CDF is:\[ F(x) = 1 - e^{-x/2}, \quad x \ge 0 \]
Step 3: We need \(P(X_1^2+X_2^2 \le 2) = F(2)\):\[ F(2) = 1 - e^{-2/2} = 1 - e^{-1} \]So \(P(X_1^2+X_2^2 \le 2) = \boxed{1-e^{-1}}\), option (C).
Step 2: A chi-square distribution with 2 degrees of freedom is the same as an Exponential distribution with mean 2 (rate \(\lambda = 1/2\)). Its CDF is:\[ F(x) = 1 - e^{-x/2}, \quad x \ge 0 \]
Step 3: We need \(P(X_1^2+X_2^2 \le 2) = F(2)\):\[ F(2) = 1 - e^{-2/2} = 1 - e^{-1} \]So \(P(X_1^2+X_2^2 \le 2) = \boxed{1-e^{-1}}\), option (C).
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