Financial Risk Manager Part 1

Explanation

For an MA(1) process, the general form is:

$y_t = \mu + \epsilon_t + \theta \epsilon_{t-1}$

Where:

• $y_t$ is today's realization
• $\mu$ is the mean (assumed to be 0 for white noise)
• $\epsilon_t$ is today's shock (0.170)
• $\epsilon_{t-1}$ is the lagged shock (-0.160)
• $\theta$ is the weight parameter (0.70)

Plugging in the values:

$y_t = 0 + 0.170 + (0.70)(-0.160)$ $y_t = 0.170 - 0.112$ $y_t = 0.058$

Wait, let me recalculate this carefully:

$y_t = \epsilon_t + \theta \epsilon_{t-1}$ $y_t = 0.170 + (0.70)(-0.160)$ $y_t = 0.170 - 0.112$ $y_t = 0.058$

However, looking at the options, 0.058 corresponds to option D, not option B (0.075). Let me double-check the calculation:

• Today's shock: 0.170
• Lagged shock: -0.160
• Theta: 0.70
• Product: 0.70 × (-0.160) = -0.112
• Sum: 0.170 + (-0.112) = 0.058

The calculation gives 0.058, which matches option D. But let me verify if there's any additional information I might be missing.

Actually, in an MA(1) process, yesterday's realization $y_{(t-1)}$ is not directly used in calculating today's realization. The MA(1) process only depends on current and lagged shocks. Therefore, the correct calculation is:

$y_t = \epsilon_t + \theta \epsilon_{t-1} = 0.170 + 0.70 \times (-0.160) = 0.170 - 0.112 = 0.058$

Therefore, the correct answer is D (0.058).

The confusion might be that yesterday's realization (0.015) was given but is not needed for the MA(1) calculation. The MA(1) process depends only on current and lagged shocks, not on previous realizations._