Question:
Let \( f_1(x) \) be the probability density function of the \( N(0,1) \) distribution, and \( f_2(x) \) be the probability density function of the \( N(0, 6) \) distribution. Let \( Y \) be a random variable with probability density function
\[f(x) = 0.6 f_1(x) + 0.4 f_2(x), \quad -\infty < x < \infty.\]
Then \( \text{Var}(Y) \) is equal to:
Let \( f_1(x) \) be the probability density function of the \( N(0,1) \) distribution, and \( f_2(x) \) be the probability density function of the \( N(0, 6) \) distribution. Let \( Y \) be a random variable with probability density function
\[f(x) = 0.6 f_1(x) + 0.4 f_2(x), \quad -\infty < x < \infty.\]
Then \( \text{Var}(Y) \) is equal to:
\[f(x) = 0.6 f_1(x) + 0.4 f_2(x), \quad -\infty < x < \infty.\]
Then \( \text{Var}(Y) \) is equal to:
Updated On: Jan 25, 2025
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The Correct Option is B
Solution and Explanation
1. Variance of a Mixture Distribution:
- For a mixture of distributions, the variance is:
\[
\mathrm{Var}(Y) = p_1\mathrm{Var}(f_1) + p_2\mathrm{Var}(f_2) + p_1p_2(\mu_1 - \mu_2)^2,
\]
where \( p_1 = 0.6, p_2 = 0.4 \), \( \mu_1 = \mu_2 = 0 \).
2. Substitute Values:
- Since \( \mu_1 = \mu_2 = 0 \), the third term vanishes:
\[
\mathrm{Var}(Y) = 0.6 \cdot 1 + 0.4 \cdot 6 = 0.6 + 2.4 = 3.
\]
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