Question

Let $Z_1$ and $Z_2$ be i.i.d. $N(0,1)$ random variables. If $Y = Z_1^2 + Z_2^2$, then $P(Y > 4)$ equals

• A. $e^{-2}$
• B. $1 - e^{-2}$
• C. $\dfrac{1}{2} e^{-2}$
• D. $e^{-4}$

Hint:

A $\chi^2$ distribution with 2 degrees of freedom is equivalent to an exponential distribution with mean 2 (rate $\frac{1}{2}$).

Answer 1

The Correct Option is A

Step 1: Recognize the distribution.
$Y = Z_1^2 + Z_2^2 \sim \chi^2(2)$, which is equivalent to an exponential distribution with parameter $\lambda = \dfrac{1}{2}$.

Step 2: Use the exponential property.
For an exponential variable with parameter $\lambda$, \[ P(Y > y) = e^{-\lambda y}. \] Here $\lambda = \dfrac{1}{2}$, $y = 4$.

Step 3: Substitute values.
\[ P(Y > 4) = e^{-(1/2) \times 4} = e^{-2}. \]

Step 4: Conclusion.
\[ \boxed{P(Y > 4) = e^{-2}}. \]