(a) Y1, Y2, ..., Yn form a random sample from a probability distribution with cumu- lative distribution function Fy(y) and probability density function fy(y). Let Y(1) = min{Y1, Y2,..., Yn}. Write the cumulative distribution function for Y(1) in terms of Fy(y) and hence show that the probability density function for Y(1) is fy, (y) = n{1– Fy(y)}"-'fy(y).

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1. (a) Y1,Y2, ... , Yn form a random sample from a probability distribution with cumu-
lative distribution function Fy (y) and probability density function fy (y). Let
Y(1) = min{Y1, Y2,..., Yn}.
Write the cumulative distribution function for Y(1) in terms of Fy(y) and hence
show that the probability density function for Y(1) is
fY, (y) = n{1 – Fy(y)}"-' fy(y).
(b) An engineering system consists of 5 components connected in series, so, if one
components fails, the system fails. The lifetimes (measured in years) of the 5
components, Yı, Y2, . . . , Y5, are all independent and identically distributed.
(i) Suppose the lifetimes follow the standard uniform distribution U (0, 1). Find
the probability density function for Y(1), the time to failure for the system,
and hence find the probability that the system functions for at least 6
months without failing.
(ii) If, instead, the lifetimes follow an exponential distribution with mean 0,
then Y(1) follows an exponential distribution with mean 0/5. Prove this
result.
Assuming that the only information available is a single observation on Y(1),
find the most powerful test of size 0.05 for Ho : 0 = 01 versus H1 : 0
where 01 < 02. (Hint: the probability density function and cumulative
distribution function for an exponential random variable with mean 0 are
f (y) = 0-1 exp(-y/0), y > 0, and F(y) = 1 – exp(-y/0), y > 0, respec-
tively.)
02,
||
Transcribed Image Text:1. (a) Y1,Y2, ... , Yn form a random sample from a probability distribution with cumu- lative distribution function Fy (y) and probability density function fy (y). Let Y(1) = min{Y1, Y2,..., Yn}. Write the cumulative distribution function for Y(1) in terms of Fy(y) and hence show that the probability density function for Y(1) is fY, (y) = n{1 – Fy(y)}"-' fy(y). (b) An engineering system consists of 5 components connected in series, so, if one components fails, the system fails. The lifetimes (measured in years) of the 5 components, Yı, Y2, . . . , Y5, are all independent and identically distributed. (i) Suppose the lifetimes follow the standard uniform distribution U (0, 1). Find the probability density function for Y(1), the time to failure for the system, and hence find the probability that the system functions for at least 6 months without failing. (ii) If, instead, the lifetimes follow an exponential distribution with mean 0, then Y(1) follows an exponential distribution with mean 0/5. Prove this result. Assuming that the only information available is a single observation on Y(1), find the most powerful test of size 0.05 for Ho : 0 = 01 versus H1 : 0 where 01 < 02. (Hint: the probability density function and cumulative distribution function for an exponential random variable with mean 0 are f (y) = 0-1 exp(-y/0), y > 0, and F(y) = 1 – exp(-y/0), y > 0, respec- tively.) 02, ||
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