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ShortTerm Actuarial Math Old Exam C Forum 

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#1




SOA 75
Im hoping the screen grab I attached shows up here. Anyway, this is the 'shifted exponential' problem.
With the substitution of y = x delta, then x=y + delta. Im good with what's inside the integral but what about the integration limits? I agree with infinity but why the lower one = 0? if before substituting it was delta, and after substituting delta = x  y, why isn't xy the correct lower limit of integration? I found a post where someone said (11 years ago) that they memorized the first and second moment equations, which I did also. However, I would like to understand where the 0 comes from if anyone has a quick answer. Thanks. 
#2




The integrals should be dx and not d theta.

#3




Then delta = xy. Not sure how that gets me to 0 for the lower integration limit though. I'm probably just being thick.

#4




x > delta, y = x  delta means y > 0.
You could get the second moment without integration as the variance plus the mean squared. The shifting does not change the variance of the exponential and increases the mean by delta. 
#5




Thanks!

#6




Quote:
Then F_Y(z) = Pr[Y <= z] = Pr[X + c <= z] = Pr[X <= z  c] = F_X(z  c). Taking the derivative with respect to z gives f_Y(z) = f_X(z  c) x 1 = f_X(z  c). That is, both the cumulative distribution function and the density function of Y look just like those for X, except they are evaluated at z  c instead of at z. Also, if X > 0, then Y > c. When you see the formula in this problem for the density of the shifted variable for z > delta, you should immediately recognize it as the density for an Exponential random variable with mean theta, except it is evaluated at z  delta instead of at z. From the above, these two things say that this X is W + delta, where W is an Exponential random variable with mean theta. Thus E[X] = E[W + delta] = theta + delta, and Var[X] = Var [W + delta] = Var[W] = theta^2.
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