2 You have three light bulbs and their lifetimes Ti are expo

(2) You have three lightbulbs, and their lfetimes T are exponential random variables with parameters (2) You have three lightbulbs, and their lifetimes T, are ulbs, and their lifetimes T, are exponential random variables with parae = i, for i = 1, 2, 3 respectively, we switch all three bulbs on at time zero. Calculate (a) the mean time you have to wait until lightbulb 3 burns out, (b) the variance of the above time, (c) P(at least one bulb is burnt out at time t), (d) the mean time you have to wait until at least one of the bulbs burns out, (e) the mean time you have to wait until all of the bulbs burn out. (f) P(exactly two bulbs are burnt out at time t) [1 mark] [1 mark] [1 mark] [1 mark] 2 marks 2 marks)

Solution

(a)

the pdf of T3 (as any Ti) is :
f3( t) = 3 e^(- 3 t)
therefore :
E( T3 ) = ( t € [0 , +oo) t * f3( t) dt
= ( t € [0 , +oo) t * 3 e^(- 3 t) dt
= (1/3) ( t € [0 , +oo) (3 t) e^(- 3 t) d(3 t) <-- the other \"3\" is here : within the \"d(3 t)\" !!
= (1/3) * 1! (use of result in above remarks)
= (1/3)

b) variance :
Var(T3) = E(T3^2) - E(T3)^2
with
E( T3^2 ) = ( t € [0 , +oo) t^2 * f3( t) dt
= ( t € [0 , +oo) t^2 * 3 e^(- 3 t) dt
= (1/3)^2 ( t € [0 , +oo) (3 t)^2 e^(- 3 t) d(3 t)
= (1/3)^2 * 2!
= 2(1/3)^2
therefore :
Var(T3) = E(T3^2) - E(T3)^2
= 2(1/3)^2 - (1/3)^2
= (1/3)^2

c)
(at least one bulb is burnt out at time t) is the complementary event to :
( T1 > t and T2 > t and T3 > t)
therefore :
P(at least one bulb is burnt out at time t) = 1 - P( T1 > t and T2 > t and T3 > t)
= 1 - P( T1 > t) * P(T2 > t) * P(T3 > t) because the Ti are independent
and, as far as the cdf are concerned, we have :
Fi( t) = P( Ti < t ) = ( t € [0 , t) i e^(- i u) du for any i
= ( t € [0 , t) e^(- i u) d(ui)
= [ e^(- i u)](between 0 and t)
= 1 - e^(- i t)
==> P( Ti > t) = e^(- i t)
therefore :
P(at least one bulb is burnt out at time t) = 1 - e^(- 1 t) e^(- 2 t) e^(- 3 t)
= 1 - e^(- [1+2+3] t) (**)

Important Remark : in fact the time Tinf where the first bulb burns out is the random variable :
Tinf = Inf( T1, T2, T3)
so we obtained in formula (**) :
P(Tinf < t ) = 1 - e^(- [1+2+3] t)
this means that the cdf of T has the form of an exponential distribution of parameter :
= 1 + 2 + 3

d)

the mean time you have to wait until at least one of the bulbs burns out :
in other words, we want : E( Tinf )
... and we already calculated in question (a) the mean time of an exponential distribution
(i.e. : T3 --> it is the inverse of the parameter !! )
therefore, we can immediately write, without calc. :
E ( Tinf ) = 1 /(1 + 2 + 3)