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    Hi, I am stuck on this question.

    The life time T of a particular brand of light bulb is modelled as follows. There is a probability p of the light-bulb blowing immediately (so that T= 0); given that the light bulb does not blow immediately, the probability of it having lifetime τ or less is 1-e^(-λτ)

    (i) Write down the cumulative distribution function F(t) of T

    I am totally confused, especially as it just says write down. Any help would be appreciated
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    (Original post by Mathstat)
    Hi, I am stuck on this question.

    The life time T of a particular brand of light bulb is modelled as follows. There is a probability p of the light-bulb blowing immediately (so that T= 0); given that the light bulb does not blow immediately, the probability of it having lifetime τ or less is 1-e^(-λτ)

    (i) Write down the cumulative distribution function F(t) of T

    I am totally confused, especially as it just says write down. Any help would be appreciated
    by definition

    F(t) </= P(T</=t)
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    (Original post by TeeEm)
    by definition

    F(t) </= P(T</=t)
    So it's just 1-e^(-λτ)?
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    (Original post by Mathstat)
    So it's just 1-e^(-λτ)?
    I have not done exponential distributions for a while but that is what I made it
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    (Original post by TeeEm)
    I have not done exponential distributions for a while but that is what I made it
    But where does the lightbulb exploding instantly with probability p come into it?
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    (Original post by Mathstat)
    But where does the lightbulb exploding instantly with probability p come into it?
    I do not what the question is asking, but I would imagine P is merely notation.

    As far as your question is concerned it makes sense
    the exponential distribution is defined between 0 and inf

    F(0) = 0
    F(inf) =1

    and by differentiation you can find the PDF , mean and all sorts.
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    (Original post by TeeEm)
    I do not what the question is asking, but I would imagine P is merely notation.

    As far as your question is concerned it makes sense
    the exponential distribution is defined between 0 and inf

    F(0) = 0
    F(inf) =1

    and by differentiation you can find the PDF , mean and all sorts.
    Is this not correct (this is working I just asked my friend for).

    P(T<τ|T=/=0)= P(T<τ union T=/=0)/P(T=/=0)

    The LHS = 1-e^(-λτ), so P(T<τ union T=/= 0) = P(T<τ)= p(1-e^(-λτ)

    Thanks for your time
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    (Original post by Mathstat)
    But where does the lightbulb exploding instantly with probability p come into it?
    I beg to differ with TeeEm (WADR). Although the question says "write down", implying no working is required, I think you do need some in this case:

    P(T\leq t) = P(T=0) + P(0 &lt; T \leq t)
    Edit: For t > 0.

    And you're told in the question P(0&lt;T\leq t\; |\; T \not= 0)

    Unusual question for A-level. Is it A-level?

    Can you take it from there?
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    (Original post by Mathstat)
    Is this not correct (this is working I just asked my friend for).

    P(T<τ|T=/=0)= P(T<τ union T=/=0)/P(T=/=0)

    The LHS = 1-e^(-λτ), so P(T<τ union T=/= 0) = P(T<τ)= p(1-e^(-λτ)

    Thanks for your time
    (Original post by ghostwalker)
    I beg to differ with TeeEm (WADR). Although the question says "write down", implying no working is required, I think you do need some in this case:

    P(T\leq t) = P(T=0) + P(0 &lt; T \leq t)

    And you're told in the question P(0&lt;T\leq t\; |\; T \not= 0)

    Unusual question for A-level. Is it A-level?

    Can you take it from there?
    I will leave you in the capable hands of ghostwalker, as stats is not my forte ...
    All the best.
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    (Original post by ghostwalker)
    I beg to differ with TeeEm (WADR). Although the question says "write down", implying no working is required, I think you do need some in this case:

    P(T\leq t) = P(T=0) + P(0 &lt; T \leq t)
    Edit: For t > 0.

    And you're told in the question P(0&lt;T\leq t\; |\; T \not= 0)

    Unusual question for A-level. Is it A-level?

    Can you take it from there?
    I am going into my second year of uni and this is on the holiday sheets.

    So is the answer (1-p)(1-e^(-λτ)) ?

    What is P(0<=T<τ union T=/=0)? Is it just 1-e^(-λτ)?

    Thanks for your help
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    (Original post by Mathstat)
    I am going into my second year of uni and this is on the holiday sheets.

    So is the answer (1-p)(1-e^(-λτ)) ?
    For t>0,

    P(T \leq t) = P(T=0) + P(0 &lt; T \leq t)

    = P(T=0) + P(0 &lt; T \leq t\; | \; T\not= 0)P(T\not=0)

    = p + (1-e^{-\lambda t})(1-p)



    What is P(0<=T<τ union T=/=0)? Is it just 1-e^(-λτ)?

    Thanks for your help
    P(0<=T<τ union T=/=0). I think you mean intersect there, rather than union.

    P(0\leq T&lt;t \cap T \not= 0) = P(0 &lt; T &lt; t)

    = P(0 &lt; T \leq t\; | \; T\not= 0)P(T\not=0)

    = (1-e^{-\lambda t})(1-p)

    I've not been consistent with the < or <= at the upper end as you've used it slightly differently, but it's not important as long as there are no discontinuities, which there aren't for t > 0.

    So you cdf will be as I previously wrote for t>0, and "p" for t=0
    Edit: The case t=0 is actually covered by the formula - ho hum.
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    (Original post by ghostwalker)
    For t>0,

    P(T \leq t) = P(T=0) + P(0 &lt; T \leq t)

    = P(T=0) + P(0 &lt; T \leq t\; | \; T\not= 0)P(T\not=0)
    For clarity, by the law of total probability that second line is really:


    = P(T=0) + P(0 &lt; T \leq t\; | \; T\not= 0)P(T\not=0)+P(0 &lt; T \leq t\; | \; T= 0)P(T=0)

    But those final terms come to 0, since T can't simultaneously be >0 and =0.

    Hence as I originally wrote.
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    (Original post by ghostwalker)
    For t>0,

    P(T \leq t) = P(T=0) + P(0 &lt; T \leq t)

    = P(T=0) + P(0 &lt; T \leq t\; | \; T\not= 0)P(T\not=0)

    = p + (1-e^{-\lambda t})(1-p)




    P(0<=T<τ union T=/=0). I think you mean intersect there, rather than union.

    P(0\leq T&lt;t \cap T \not= 0) = P(0 &lt; T &lt; t)

    = P(0 &lt; T \leq t\; | \; T\not= 0)P(T\not=0)

    = (1-e^{-\lambda t})(1-p)

    I've not been consistent with the < or <= at the upper end as you've used it slightly differently, but it's not important as long as there are no discontinuities, which there aren't for t > 0.

    So you cdf will be as I previously wrote for t>0, and "p" for t=0
    Thanks you so much . All my friends were stuck on this as well.
 
 
 
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