Compute the inverse cdfs of the following density functions and describe how to generate
random variables from these distributions based on the inverse-cdf method. f(x) = 1 / 2b e -|x|/b,
b > 0, x R. (1) f(x) = k(x - a) k-1 / (b - a)k, k > 0, a x b. (2).
Solution
I will use I[a,b] for the integral from a to b and E[a,b] for evaluated from a to b
For - infinity < x < 0, F(x) = |[-infinity,x] 1/2b e ^ x/b = 1/2 e ^x/b E[-infinity, x] =
1/2 e ^x/b - 0 = 1/2 e ^x/b
For 0 < x < infinity, as the value of integral above = 1/2 at 0, F(x) =
1/2 + |[0,x] 1/2b e^-x/b = 1/2 - 1/2e^-x/bE[0, x] =
1/2 - 1/2 e^-x/b - -1/2 = 1 - 1/2e^-x/b
Then, if 1/2 e^x/b = Y, e^x/b = 2Y, and x/b = ln(2Y) and x = ln(2Y) b Y <= 1/2
if1 - 1/2e^-x/b = Y
1 - Y = 1/2 e^-x/b
2 - 2Y = e^-x/b
ln(2-2Y) = -x/b
x = -ln(2-2Y) b Y >= 1/2
Our solution is
x = ln(2Y) b Y <= 1/2
x = -ln(2-2Y) b Y >= 1/2
where Y ~ U[0,1]
As f(x) = k(x-a)^(k-1)/(b-a)^k a <= x <= b
F(x) = I[a,x] k(x-a)^(k-1)/(b-a)^k =(x-a)^k/(b-a)^k E[a,x] =
(x-a)^k/(b-a)^k - 0 =
(x-a)^k/(b-a)^k
Then, if(x-a)^k/(b-a)^k = y
(x-a)^k =(b-a)^k y
Taking the k\'th root on both sides
x-a = (b-a) y^(1/k)
x = a +(b-a) y^(1/k)
.
Compute the inverse cdfs of the following density functions and descr.pdf
1. Compute the inverse cdfs of the following density functions and describe how to generate
random variables from these distributions based on the inverse-cdf method. f(x) = 1 / 2b e -|x|/b,
b > 0, x R. (1) f(x) = k(x - a) k-1 / (b - a)k, k > 0, a x b. (2).
Solution
I will use I[a,b] for the integral from a to b and E[a,b] for evaluated from a to b
For - infinity < x < 0, F(x) = |[-infinity,x] 1/2b e ^ x/b = 1/2 e ^x/b E[-infinity, x] =
1/2 e ^x/b - 0 = 1/2 e ^x/b
For 0 < x < infinity, as the value of integral above = 1/2 at 0, F(x) =
1/2 + |[0,x] 1/2b e^-x/b = 1/2 - 1/2e^-x/bE[0, x] =
1/2 - 1/2 e^-x/b - -1/2 = 1 - 1/2e^-x/b
Then, if 1/2 e^x/b = Y, e^x/b = 2Y, and x/b = ln(2Y) and x = ln(2Y) b Y <= 1/2
if1 - 1/2e^-x/b = Y
1 - Y = 1/2 e^-x/b
2 - 2Y = e^-x/b
ln(2-2Y) = -x/b
x = -ln(2-2Y) b Y >= 1/2
Our solution is
x = ln(2Y) b Y <= 1/2
x = -ln(2-2Y) b Y >= 1/2
2. where Y ~ U[0,1]
As f(x) = k(x-a)^(k-1)/(b-a)^k a <= x <= b
F(x) = I[a,x] k(x-a)^(k-1)/(b-a)^k =(x-a)^k/(b-a)^k E[a,x] =
(x-a)^k/(b-a)^k - 0 =
(x-a)^k/(b-a)^k
Then, if(x-a)^k/(b-a)^k = y
(x-a)^k =(b-a)^k y
Taking the k'th root on both sides
x-a = (b-a) y^(1/k)
x = a +(b-a) y^(1/k)
![Compute the inverse cdfs of the following density functions and describe how to generate
random variables from these distributions based on the inverse-cdf method. f(x) = 1 / 2b e -|x|/b,
b > 0, x R. (1) f(x) = k(x - a) k-1 / (b - a)k, k > 0, a x b. (2).
Solution
I will use I[a,b] for the integral from a to b and E[a,b] for evaluated from a to b
For - infinity < x < 0, F(x) = |[-infinity,x] 1/2b e ^ x/b = 1/2 e ^x/b E[-infinity, x] =
1/2 e ^x/b - 0 = 1/2 e ^x/b
For 0 < x < infinity, as the value of integral above = 1/2 at 0, F(x) =
1/2 + |[0,x] 1/2b e^-x/b = 1/2 - 1/2e^-x/bE[0, x] =
1/2 - 1/2 e^-x/b - -1/2 = 1 - 1/2e^-x/b
Then, if 1/2 e^x/b = Y, e^x/b = 2Y, and x/b = ln(2Y) and x = ln(2Y) b Y <= 1/2
if1 - 1/2e^-x/b = Y
1 - Y = 1/2 e^-x/b
2 - 2Y = e^-x/b
ln(2-2Y) = -x/b
x = -ln(2-2Y) b Y >= 1/2
Our solution is
x = ln(2Y) b Y <= 1/2
x = -ln(2-2Y) b Y >= 1/2](https://image.slidesharecdn.com/computetheinversecdfsofthefollowingdensityfunctionsanddescr-230325040514-df7e568c/85/Compute-the-inverse-cdfs-of-the-following-density-functions-and-descr-pdf-1-320.jpg)
![where Y ~ U[0,1]
As f(x) = k(x-a)^(k-1)/(b-a)^k a <= x <= b
F(x) = I[a,x] k(x-a)^(k-1)/(b-a)^k =(x-a)^k/(b-a)^k E[a,x] =
(x-a)^k/(b-a)^k - 0 =
(x-a)^k/(b-a)^k
Then, if(x-a)^k/(b-a)^k = y
(x-a)^k =(b-a)^k y
Taking the k'th root on both sides
x-a = (b-a) y^(1/k)
x = a +(b-a) y^(1/k)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)