f(y)= (1/c)e^(-y/2) if y is greater than or equal to 0
and f(y)=(1/c)e^(y/2) if y is less than 0
a) What is c?
b)What is the probability distribution function?
To find the value of c, we can use the fact that the total probability of a probability distribution function (PDF) must be equal to 1.
For y ≥ 0, the PDF is given by f(y) = (1/c)e^(-y/2).
For y < 0, the PDF is given by f(y) = (1/c)e^(y/2).
To find c, let's integrate the PDF over the entire range:
For y ≥ 0: ∫[0,∞] (1/c)e^(-y/2) dy
For y < 0: ∫[-∞,0] (1/c)e^(y/2) dy
To integrate the first part, we can use the substitution u = -y/2, which implies du = -dy/2.
For y ≥ 0: ∫[0,∞] (1/c)e^(-y/2) dy = ∫[0,∞] (1/c)e^u (-du/2)
= (-1/2c) ∫[0,∞] e^u du
= (-1/2c) [e^u] [0,∞]
= (-1/2c) (e^∞ - e^0)
= (-1/2c) (1 - 1)
= 0
Since this part of the integral evaluates to 0, it means that all the probability lies in the second part, where y < 0. We can set this part equal to 1 to find c.
For y < 0: ∫[-∞, 0] (1/c)e^(y/2) dy = ∫[-∞, 0] (1/c)e^(y/2) dy = 1
So, we have (1/c) ∫[-∞,0] e^(y/2) dy = 1.
By integrating, we find (2/c) (e^(y/2)) [ -∞, 0] = 1.
Using the limits, we have (2/c) (e^(0/2) - e^(-∞/2)) = 1.
Simplifying further, we get (2/c) (1 - 0) = 1, which gives 2/c = 1.
Therefore, c = 2.
a) The value of c is 2.
Now, let's determine the probability distribution function (PDF).
Given that c = 2, we can rewrite the PDF as follows:
For y ≥ 0: f(y) = (1/2)e^(-y/2)
For y < 0: f(y) = (1/2)e^(y/2)
b) The probability distribution function (PDF) is:
f(y) = (1/2)e^(-|y|/2)
The absolute value |y| ensures that the function is symmetric around the y-axis and smoothly transitions as y changes from positive to negative values.