Finding the distribution function of a continuous random variable with a density function including a minimum
I have been scratching my head about this question for a long time.
I found one other question on here that included a minimum function for probability functions but unfortunately there wasn't enough information to make me understand this enough to proceed.
The question states:
Let X is a continuous random variable with probability density function
$$f(x) = c min left( 1, dfrac 1 {x^4} right)$$
(a) Find c and the (cumulative) distribution function of X.
(b) Find EX, VarX, and the median of X.
Based on the example I saw, this gave me a density function:
$$f(x) = begin{cases}
c & mbox{for } -1 ≤ x ≤ 1\
tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
Is this correct?
If so then by setting the summation of the integrals of these f(x) cases gives me a value of 3/8 for c.
And if so would I be correct to get a distribution function as:
$$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$
I'm not sure about this though and even if this is correct, I'm not sure how to use this for part b.
probability-distributions
asked Nov 26 at 18:07
Mike Campbell
62
|
show 2 more comments
I have been scratching my head about this question for a long time.
I found one other question on here that included a minimum function for probability functions but unfortunately there wasn't enough information to make me understand this enough to proceed.
The question states:
Let X is a continuous random variable with probability density function
$$f(x) = c min left( 1, dfrac 1 {x^4} right)$$
(a) Find c and the (cumulative) distribution function of X.
(b) Find EX, VarX, and the median of X.
Based on the example I saw, this gave me a density function:
$$f(x) = begin{cases}
c & mbox{for } -1 ≤ x ≤ 1\
tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
Is this correct?
If so then by setting the summation of the integrals of these f(x) cases gives me a value of 3/8 for c.
And if so would I be correct to get a distribution function as:
$$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$
I'm not sure about this though and even if this is correct, I'm not sure how to use this for part b.
probability-distributions
asked Nov 26 at 18:07
Mike Campbell
62
Have you noticed that, with your definition, $F(x)$ is piecewise constant?
– Roberto Rastapopoulos
Nov 26 at 18:10
I feel like the problem for me lies with my original definition for f(x). It is the min part of the function that has me confused.
– Mike Campbell
Nov 26 at 18:14
Your alternative definition of $f$ is correct. :)
– Roberto Rastapopoulos
Nov 26 at 18:19
so this one is correct?: $$f(x) = begin{cases} c & mbox{for } -1 ≤ x ≤ 1\ tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
– Mike Campbell
Nov 26 at 18:20
yes, that definition is correct.
– Roberto Rastapopoulos
Nov 26 at 18:21
|
show 2 more comments
I have been scratching my head about this question for a long time.
I found one other question on here that included a minimum function for probability functions but unfortunately there wasn't enough information to make me understand this enough to proceed.
The question states:
Let X is a continuous random variable with probability density function
$$f(x) = c min left( 1, dfrac 1 {x^4} right)$$
(a) Find c and the (cumulative) distribution function of X.
(b) Find EX, VarX, and the median of X.
Based on the example I saw, this gave me a density function:
$$f(x) = begin{cases}
c & mbox{for } -1 ≤ x ≤ 1\
tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
Is this correct?
If so then by setting the summation of the integrals of these f(x) cases gives me a value of 3/8 for c.
And if so would I be correct to get a distribution function as:
$$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$
I'm not sure about this though and even if this is correct, I'm not sure how to use this for part b.
probability-distributions
asked Nov 26 at 18:07
Mike Campbell
62
I have been scratching my head about this question for a long time.
I found one other question on here that included a minimum function for probability functions but unfortunately there wasn't enough information to make me understand this enough to proceed.
The question states:
Let X is a continuous random variable with probability density function
$$f(x) = c min left( 1, dfrac 1 {x^4} right)$$
(a) Find c and the (cumulative) distribution function of X.
(b) Find EX, VarX, and the median of X.
Based on the example I saw, this gave me a density function:
$$f(x) = begin{cases}
c & mbox{for } -1 ≤ x ≤ 1\
tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
Is this correct?
If so then by setting the summation of the integrals of these f(x) cases gives me a value of 3/8 for c.
And if so would I be correct to get a distribution function as:
$$F(x) = begin{cases} int_{1}^{infty} tfrac{3}{8x^4} dx & text{if } x>1 \ int_{-1}^{1} tfrac{3}{8} dx & text{if } -1 ≤ x ≤ 1 \ int_{-infty}^{-1} tfrac{3}{8x^4} & text{if } x < -1end{cases}$$
I'm not sure about this though and even if this is correct, I'm not sure how to use this for part b.
probability-distributions
probability-distributions
asked Nov 26 at 18:07
Mike Campbell
62
asked Nov 26 at 18:07
Mike Campbell
62
asked Nov 26 at 18:07
Mike Campbell
62
asked Nov 26 at 18:07
Mike Campbell
62
asked Nov 26 at 18:07
Mike Campbell
62
62
Have you noticed that, with your definition, $F(x)$ is piecewise constant?
– Roberto Rastapopoulos
Nov 26 at 18:10
I feel like the problem for me lies with my original definition for f(x). It is the min part of the function that has me confused.
– Mike Campbell
Nov 26 at 18:14
Your alternative definition of $f$ is correct. :)
– Roberto Rastapopoulos
Nov 26 at 18:19
so this one is correct?: $$f(x) = begin{cases} c & mbox{for } -1 ≤ x ≤ 1\ tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
– Mike Campbell
Nov 26 at 18:20
yes, that definition is correct.
– Roberto Rastapopoulos
Nov 26 at 18:21
|
show 2 more comments
Have you noticed that, with your definition, $F(x)$ is piecewise constant?
– Roberto Rastapopoulos
Nov 26 at 18:10
I feel like the problem for me lies with my original definition for f(x). It is the min part of the function that has me confused.
– Mike Campbell
Nov 26 at 18:14
Your alternative definition of $f$ is correct. :)
– Roberto Rastapopoulos
Nov 26 at 18:19
so this one is correct?: $$f(x) = begin{cases} c & mbox{for } -1 ≤ x ≤ 1\ tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
– Mike Campbell
Nov 26 at 18:20
yes, that definition is correct.
– Roberto Rastapopoulos
Nov 26 at 18:21
Have you noticed that, with your definition, $F(x)$ is piecewise constant?
– Roberto Rastapopoulos
Nov 26 at 18:10
Have you noticed that, with your definition, $F(x)$ is piecewise constant?
– Roberto Rastapopoulos
Nov 26 at 18:10
I feel like the problem for me lies with my original definition for f(x). It is the min part of the function that has me confused.
– Mike Campbell
Nov 26 at 18:14
I feel like the problem for me lies with my original definition for f(x). It is the min part of the function that has me confused.
– Mike Campbell
Nov 26 at 18:14
Your alternative definition of $f$ is correct. :)
– Roberto Rastapopoulos
Nov 26 at 18:19
Your alternative definition of $f$ is correct. :)
– Roberto Rastapopoulos
Nov 26 at 18:19
so this one is correct?: $$f(x) = begin{cases} c & mbox{for } -1 ≤ x ≤ 1\ tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
– Mike Campbell
Nov 26 at 18:20
so this one is correct?: $$f(x) = begin{cases} c & mbox{for } -1 ≤ x ≤ 1\ tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
– Mike Campbell
Nov 26 at 18:20
yes, that definition is correct.
– Roberto Rastapopoulos
Nov 26 at 18:21
yes, that definition is correct.
– Roberto Rastapopoulos
Nov 26 at 18:21
|
show 2 more comments
1 Answer
1
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oldest
votes
By definition,
$$F(x) = int_{-infty}^x f(y) , dy, $$
so we obtain:
$$F(x) = begin{cases}
int_{-infty}^{x} tfrac{3}{8y^4} dy = frac{1}{8} , |x|^{-3}& text{if } x < -1\
F(-1) + int_{-1}^{x} tfrac{3}{8} dy = frac{1}{8} (4 + 3x) & text{if } -1 ≤ x ≤ 1 \
F(1) + int_{1}^{x} tfrac{3}{8y^4}dx = 1 - frac{1}{8} |x|^{-3} & text{if } x>1 end{cases}
$$
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
add a comment |
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1 Answer
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oldest
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By definition,
$$F(x) = int_{-infty}^x f(y) , dy, $$
so we obtain:
$$F(x) = begin{cases}
int_{-infty}^{x} tfrac{3}{8y^4} dy = frac{1}{8} , |x|^{-3}& text{if } x < -1\
F(-1) + int_{-1}^{x} tfrac{3}{8} dy = frac{1}{8} (4 + 3x) & text{if } -1 ≤ x ≤ 1 \
F(1) + int_{1}^{x} tfrac{3}{8y^4}dx = 1 - frac{1}{8} |x|^{-3} & text{if } x>1 end{cases}
$$
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
add a comment |
By definition,
$$F(x) = int_{-infty}^x f(y) , dy, $$
so we obtain:
$$F(x) = begin{cases}
int_{-infty}^{x} tfrac{3}{8y^4} dy = frac{1}{8} , |x|^{-3}& text{if } x < -1\
F(-1) + int_{-1}^{x} tfrac{3}{8} dy = frac{1}{8} (4 + 3x) & text{if } -1 ≤ x ≤ 1 \
F(1) + int_{1}^{x} tfrac{3}{8y^4}dx = 1 - frac{1}{8} |x|^{-3} & text{if } x>1 end{cases}
$$
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
add a comment |
By definition,
$$F(x) = int_{-infty}^x f(y) , dy, $$
so we obtain:
$$F(x) = begin{cases}
int_{-infty}^{x} tfrac{3}{8y^4} dy = frac{1}{8} , |x|^{-3}& text{if } x < -1\
F(-1) + int_{-1}^{x} tfrac{3}{8} dy = frac{1}{8} (4 + 3x) & text{if } -1 ≤ x ≤ 1 \
F(1) + int_{1}^{x} tfrac{3}{8y^4}dx = 1 - frac{1}{8} |x|^{-3} & text{if } x>1 end{cases}
$$
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
By definition,
$$F(x) = int_{-infty}^x f(y) , dy, $$
so we obtain:
$$F(x) = begin{cases}
int_{-infty}^{x} tfrac{3}{8y^4} dy = frac{1}{8} , |x|^{-3}& text{if } x < -1\
F(-1) + int_{-1}^{x} tfrac{3}{8} dy = frac{1}{8} (4 + 3x) & text{if } -1 ≤ x ≤ 1 \
F(1) + int_{1}^{x} tfrac{3}{8y^4}dx = 1 - frac{1}{8} |x|^{-3} & text{if } x>1 end{cases}
$$
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
edited Nov 26 at 18:28
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
answered Nov 26 at 18:22
Roberto Rastapopoulos
869424
869424
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
add a comment |
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
Ah I see, so I missed out the F(1) and F(-1)
– Mike Campbell
Nov 26 at 18:27
add a comment |
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Have you noticed that, with your definition, $F(x)$ is piecewise constant?
– Roberto Rastapopoulos
Nov 26 at 18:10
I feel like the problem for me lies with my original definition for f(x). It is the min part of the function that has me confused.
– Mike Campbell
Nov 26 at 18:14
Your alternative definition of $f$ is correct. :)
– Roberto Rastapopoulos
Nov 26 at 18:19
so this one is correct?: $$f(x) = begin{cases} c & mbox{for } -1 ≤ x ≤ 1\ tfrac{c}{x^4} & mbox{for } |x| > 1end{cases}$$
– Mike Campbell
Nov 26 at 18:20
yes, that definition is correct.
– Roberto Rastapopoulos
Nov 26 at 18:21