Financial Risk Manager Part 1

Financial Risk Manager Part 1

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Let X have the following probability density function:

fX(x)={0.15x=10.25x=20.35x=3Cx=4f_X(x) = \begin{cases} 0.15 & x = 1 \\ 0.25 & x = 2 \\ 0.35 & x = 3 \\ C & x = 4 \end{cases}

Calculate the mode of the distribution._

TTanishq



Explanation:

Solution

First, we need to find the value of C. Since this is a probability density function, the sum of all probabilities must equal 1:

C=1βˆ’0.15βˆ’0.25βˆ’0.35=0.25C = 1 - 0.15 - 0.25 - 0.35 = 0.25

Now we have the complete probability distribution:

  • P(X=1) = 0.15
  • P(X=2) = 0.25
  • P(X=3) = 0.35
  • P(X=4) = 0.25

The mode is the value with the highest probability in the distribution. Comparing the probabilities:

  • P(X=1) = 0.15
  • P(X=2) = 0.25
  • P(X=3) = 0.35 (highest)
  • P(X=4) = 0.25

Therefore, the mode is x = 3, which corresponds to option D.

Key Concept: The mode of a probability distribution is the value that occurs with the highest probability.

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