Pdf and Coefficient of Kurtosis

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SUMMARY

The discussion focuses on finding the probability density function (pdf) and the coefficient of kurtosis for the random variable W, defined as W = X - Y, where X and Y are independent exponential random variables with a common rate parameter λ. The derived pdf is f(w) = (λ/2)e^(λw) for -∞ < w < 0 and f(w) = (λ/2)e^(-λw) for 0 < w < ∞. The coefficient of kurtosis is calculated using the formula K = E(((W - μ)/σ)^4) - 3, where the mean (μ) and standard deviation (σ) are derived separately for the two regions of W, but the calculation must consider the entire range from -∞ to +∞.

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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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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!
 

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