Given the cdf of a r.v X F(x)=1-e^-x for x >= 0 Calculate (a)variance, (b) skewn

Question

Given the cdf of a r.v X F(x)=1-e^-x for x >= 0
Calculate (a)variance, (b) skewness Derive its MGF, find 1st, 2nd, 3rd derivative of the MGF Derive its natural log function of the MGF (lnMx(t)), and find its 1st, 2nd, and 3rd derivative.
Show that mean=[lnMx(t)]' t=0 and var = [lnMx(t)]'' t=0.
Given the cdf of a r.v X F(x)=1-e^-x for x >= 0
Calculate (a)variance, (b) skewness Derive its MGF, find 1st, 2nd, 3rd derivative of the MGF Derive its natural log function of the MGF (lnMx(t)), and find its 1st, 2nd, and 3rd derivative.
Show that mean=[lnMx(t)]' t=0 and var = [lnMx(t)]'' t=0.
F(x)=1-e^-x for x >= 0
Calculate (a)variance, (b) skewness Derive its MGF, find 1st, 2nd, 3rd derivative of the MGF Derive its natural log function of the MGF (lnMx(t)), and find its 1st, 2nd, and 3rd derivative.
Show that mean=[lnMx(t)]' t=0 and var = [lnMx(t)]'' t=0.

Explanation / Answer

pdf is

F'(x) = e-x,x>0

This is exponential distibution with parameter 1.

Variance = 1/1^1 =1

b)

Median = ln 2/1 <1

Hence median falls to the left of mean

Rightly skewed

MGF = 1/(1-t) for t<1

= 1+t+t2+t3+...

ln MGF= -ln(1-t)

= -t-t^2/2 -t^3/3-t^4/4.....

I derivative =-1-t-t^2-t^3..... = -MGF

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