This Exam P sample reference tests Uniform Distribution. The expected amount above the threshold is 180: the threshold is exceeded with probability 0.60 and the conditional mean excess is 300. Therefore the payment mean is 180p=90, giving p=0.50 and choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis divides 90 by the maximum possible excess 1000-400=600 and obtains 0.15. A maximum is not an expected excess.
BThis divides 90 by the unconditional mean loss 500 and obtains 0.18, ignoring both the threshold and the positive-part payment.
CThis divides by the conditional mean excess 300 and obtains 0.30, but omits the 0.60 probability that the threshold is exceeded.
EUsing the nonexceedance probability 0.40 instead of the exceedance probability 0.60 gives an erroneous expected excess 0.40(300)=120 and 90/120=0.75.
Original practice · fully worked
Original variant: calibrate an overflow threshold
A solar generator's hourly output X is exponentially distributed with mean 10 units. A controller diverts one half of any output above a threshold c, so the diverted amount is D=0.5(X-c)+. The mean diverted amount is 2 units. Calculate c.
A 2.000
B 5.000
C 6.931
D 9.163
E 10.000
Variant answer in brief
For an exponential output with mean 10, E[(X-c)+]=10exp(-c/10). Thus 5exp(-c/10)=2 and c=10ln(2.5)=9.163, so choice D is correct.
Setup
Setup
Use the tail-integral formula for the expected positive-part excess.
E[(X−c)+]=∫c∞Pr(X>t)dt
Pr(X>t)=e−t/10
Model
Model
Integrate the exponential tail and apply the one-half diversion fraction.
E[(X−c)+]=10e−c/10
E[D]=5e−c/10
Compute
Compute
Set the diverted mean equal to two and invert the exponential.
5e−c/10=2
e−c/10=0.40
c=−10ln(0.40)=10ln(2.5)=9.162907319…
Answer
Answer
The overflow threshold is approximately 9.163 units.
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