Financial Risk Manager Part 1

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Given the following joint probability density function of two random variables:

f(x_1, x_2) = \begin{cases} 4x_1x_2, & 0 < x_1 < 1, \, 0 < x_2 < 1 \\ 0, & \text{elsewhere} \end{cases}

Find $E(X_1 | X_2 = 0.5)$

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TTanishq

0.01

0.54

0.33

0.67

Explanation:

Step-by-Step Solution

The conditional distribution is given by:

f_{(X_1|X_2)}(x_1 | X_2 = x_2) = \frac{f_{(X_1,X_2)}(x_1, x_2)}{f_{(X_2)}(x_2)}

So,

f(x_1 | X_2 = 0.5) = \frac{f_{(x_1,x_2)}(x_1, x_2 = 0.5)}{f_{(x_2)}(x_2 = 0.5)}

The marginal distribution of $X_2$ is:

f_{(x_2)}(x_2) = \int_{-\infty}^{\infty} f_{(x_1,x_2)}(x_1, x_2) \, dx_1 = \int_0^1 4x_1x_2 \, dx_1 = \left[2x_1^2x_2\right]_0^1 = 2x_2 - 0 = 2x_2

So, $f_{(x_2)}(x_2 = 0.5) = 2 \times 0.5 = 1$

f(x_1 | X_2 = 0.5) = \frac{f_{(x_1,x_2)}(x_1, x_2 = 0.5)}{f_{(x_2)}(x_2 = 0.5)} = \frac{4x_1 \times 0.5}{1} = 2x_1

Therefore, the conditional density is $f(x_1 | X_2 = 0.5) = 2x_1$ for $0 < x_1 < 1$ .

E(X_1 | X_2 = 0.5) = \int_0^1 x_1 \cdot f(x_1 | X_2 = 0.5) \, dx_1 = \int_0^1 x_1 \cdot 2x_1 \, dx_1 = 2 \int_0^1 x_1^2 \, dx_1 = 2 \left[\frac{x_1^3}{3}\right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3} \approx 0.67

✅ Final Answer: D. 0.67_

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