Pdf and Coefficient of Kurtosis

In summary: So, the integral will be ∫ _{-∞}^∞ ....., whose integrand will be different in the two regions, and the integral will be a sum of two integrals, one for the negative region and one for the positive region. That will lead to a very nice answer, but the integrand is a bit of a mess. In summary, we found the pdf of W by breaking it up into two parts: (f(w) = (λ/2)e^(λw) for w < 0 and f(w) = (λ/2)e^(-λw) for w > 0). We then found the coefficient of kurtosis by breaking it up into two regions (-
  • #1
Aria1
21
0

Homework Statement



Let X and Y be two independent exponential random variables with a common rate parameter λ>0. Let W=X-Y.
a)Find the pdf of W=X-Y
b) Find the coefficient of kurtosis of W=X-Y

Homework Equations



f(x)=λe^(-λx)
f(y)=λe^(-λy)
f(x,y) = (λ^2)e^(-λ(x+y))
Kurtosis = E(((W-μ)/σ)^4)-3

The Attempt at a Solution


I first attempted to find the cdf of W, broken up into two parts: -∞<w<0 and 0<w<∞.
-∞<w<0 : ∫from -w to ∞ ∫from 0 to y+w (f(x,y))dxdy = (1/2)e^(λw)
0<w<∞ : ∫from 0 to ∞ ∫from 0 to y+w (f(x,y))dxdy = -(1/2)e^(-λw) + 1

I then derived the cdf to get a pdf of
f(w) = (λ/2)e^(λw) , -∞<w<0
(λ/2)e^(-λw) , 0<w<∞

For the coefficient of kurtosis, I kept the problem broken into the two regions of W.
For -∞<w<0:
μ= -1/2λ and σ= √(3)/2λ
Kurtosis = ∫from -∞ to 0 (((2λw+1)^4)/9)*((λ/2)e^(λw))dw - 3
For 0<w<∞:
μ= 1/2λ and σ= √(3)/2λ
Kurtosis = ∫from 0 to ∞ (((2λw-1)^4)/9)*((λ/2)e^(-λw))dw - 3


However, when I tried to calculate in mathematica, the program could not complete it. I have no idea what's wrong or what part of the problem needs to be corrected, but if someone could please look at it and let me know, that would be wonderful!
Thank you in advance!
 
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  • #2
Aria1 said:

Homework Statement



Let X and Y be two independent exponential random variables with a common rate parameter λ>0. Let W=X-Y.
a)Find the pdf of W=X-Y
b) Find the coefficient of kurtosis of W=X-Y

Homework Equations



f(x)=λe^(-λx)
f(y)=λe^(-λy)
f(x,y) = (λ^2)e^(-λ(x+y))
Kurtosis = E(((W-μ)/σ)^4)-3

The Attempt at a Solution


I first attempted to find the cdf of W, broken up into two parts: -∞<w<0 and 0<w<∞.
-∞<w<0 : ∫from -w to ∞ ∫from 0 to y+w (f(x,y))dxdy = (1/2)e^(λw)
0<w<∞ : ∫from 0 to ∞ ∫from 0 to y+w (f(x,y))dxdy = -(1/2)e^(-λw) + 1

I then derived the cdf to get a pdf of
f(w) = (λ/2)e^(λw) , -∞<w<0
(λ/2)e^(-λw) , 0<w<∞

For the coefficient of kurtosis, I kept the problem broken into the two regions of W.
For -∞<w<0:
μ= -1/2λ and σ= √(3)/2λ
Kurtosis = ∫from -∞ to 0 (((2λw+1)^4)/9)*((λ/2)e^(λw))dw - 3
For 0<w<∞:
μ= 1/2λ and σ= √(3)/2λ
Kurtosis = ∫from 0 to ∞ (((2λw-1)^4)/9)*((λ/2)e^(-λw))dw - 3


However, when I tried to calculate in mathematica, the program could not complete it. I have no idea what's wrong or what part of the problem needs to be corrected, but if someone could please look at it and let me know, that would be wonderful!
Thank you in advance!

The Kurtosis is ##E(W - EW)^4/\sigma^4 \;- \; 3##, and the density function f(w) of W is symmetric about w = 0, so EW = 0!

It is a very bad mistake to compute means of the separate (w > 0) and (w < 0) parts; W just has one part, and it goes from -∞ to +∞. The density function f(w) has a different formula in the two regions {w>0} and {w<0}, but that is a separate issue!
 

Related to Pdf and Coefficient of Kurtosis

What is a PDF and how does it relate to the coefficient of kurtosis?

A PDF, or probability density function, is a mathematical function that describes the probability of a continuous random variable taking on a certain value. The coefficient of kurtosis is a measure of the "peakedness" or "flatness" of a distribution, and it is often used to describe the shape of a PDF.

How is the coefficient of kurtosis calculated?

The coefficient of kurtosis is calculated by taking the fourth standardized moment of a distribution and dividing it by the square of the second standardized moment (also known as the variance). This value is then subtracted by 3 to obtain the final coefficient of kurtosis.

What values of the coefficient of kurtosis indicate a "normal" distribution?

A coefficient of kurtosis of 3 is typically used as the benchmark for a "normal" distribution. Distributions with a coefficient of kurtosis less than 3 are considered "platykurtic" or "flat," while those with a coefficient greater than 3 are considered "leptokurtic" or "peaked."

How does the coefficient of kurtosis affect the shape of a distribution?

The coefficient of kurtosis provides information about the shape of a distribution. A high coefficient indicates that the distribution has a relatively high peak and heavy tails, while a low coefficient indicates a flatter, more spread out distribution. A coefficient of 3 indicates a symmetrical, "normal" distribution.

What are the applications of the coefficient of kurtosis in research and analysis?

The coefficient of kurtosis is commonly used in finance, statistics, and other fields to describe the shape of a distribution and identify any outliers or deviations from a normal distribution. It can also be used to compare the kurtosis of different datasets or to identify potential issues with data quality or measurement errors.

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