Laplace transform of exponential distributed random varibale











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Let $Tsim Exp(lambda)$. I want to calculate the laplace transform
$$E[e^{-delta T}]$$



So far I get $$E[e^{-delta T}]=int_0^infty e^{-delta x}lambda e^{-lambda x}dx=lambdaint_0^infty e^{-(delta+lambda) x}dx=lambdacdotbigg[-frac{1}{delta+lambda}e^{-(delta+lambda)x}bigg]_0^infty=frac{lambda}{delta+lambda}$$



Is this right?










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    Search for "moment generating function of exponential distribution".
    – StubbornAtom
    10 hours ago










  • Why you don't just write yes!?
    – Stockfish
    10 hours ago










  • His comment was before my edit. ;-) So I consider it being right now, thank you.
    – user408858
    10 hours ago

















up vote
0
down vote

favorite












Let $Tsim Exp(lambda)$. I want to calculate the laplace transform
$$E[e^{-delta T}]$$



So far I get $$E[e^{-delta T}]=int_0^infty e^{-delta x}lambda e^{-lambda x}dx=lambdaint_0^infty e^{-(delta+lambda) x}dx=lambdacdotbigg[-frac{1}{delta+lambda}e^{-(delta+lambda)x}bigg]_0^infty=frac{lambda}{delta+lambda}$$



Is this right?










share|cite|improve this question




















  • 1




    Search for "moment generating function of exponential distribution".
    – StubbornAtom
    10 hours ago










  • Why you don't just write yes!?
    – Stockfish
    10 hours ago










  • His comment was before my edit. ;-) So I consider it being right now, thank you.
    – user408858
    10 hours ago















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $Tsim Exp(lambda)$. I want to calculate the laplace transform
$$E[e^{-delta T}]$$



So far I get $$E[e^{-delta T}]=int_0^infty e^{-delta x}lambda e^{-lambda x}dx=lambdaint_0^infty e^{-(delta+lambda) x}dx=lambdacdotbigg[-frac{1}{delta+lambda}e^{-(delta+lambda)x}bigg]_0^infty=frac{lambda}{delta+lambda}$$



Is this right?










share|cite|improve this question















Let $Tsim Exp(lambda)$. I want to calculate the laplace transform
$$E[e^{-delta T}]$$



So far I get $$E[e^{-delta T}]=int_0^infty e^{-delta x}lambda e^{-lambda x}dx=lambdaint_0^infty e^{-(delta+lambda) x}dx=lambdacdotbigg[-frac{1}{delta+lambda}e^{-(delta+lambda)x}bigg]_0^infty=frac{lambda}{delta+lambda}$$



Is this right?







probability functional-analysis probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago

























asked 10 hours ago









user408858

15810




15810








  • 1




    Search for "moment generating function of exponential distribution".
    – StubbornAtom
    10 hours ago










  • Why you don't just write yes!?
    – Stockfish
    10 hours ago










  • His comment was before my edit. ;-) So I consider it being right now, thank you.
    – user408858
    10 hours ago
















  • 1




    Search for "moment generating function of exponential distribution".
    – StubbornAtom
    10 hours ago










  • Why you don't just write yes!?
    – Stockfish
    10 hours ago










  • His comment was before my edit. ;-) So I consider it being right now, thank you.
    – user408858
    10 hours ago










1




1




Search for "moment generating function of exponential distribution".
– StubbornAtom
10 hours ago




Search for "moment generating function of exponential distribution".
– StubbornAtom
10 hours ago












Why you don't just write yes!?
– Stockfish
10 hours ago




Why you don't just write yes!?
– Stockfish
10 hours ago












His comment was before my edit. ;-) So I consider it being right now, thank you.
– user408858
10 hours ago






His comment was before my edit. ;-) So I consider it being right now, thank you.
– user408858
10 hours ago

















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