The life of a machine ($x$) has an exponential distribution with $lambda = 1$. What is the expected life of...
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Could someone please explain the solution:
Solution:
$E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
$F_X(x) = P(X leq x)$
$F_{X | X > 1}(x | x > 1) = P(X leq x | x > 1) = frac{P(X leq x, X > 1)}{P(X > 1)} = frac{F_{X}(x) - F_{X}(1)}{1 - F_{X}(1)}$
$= frac{F_{X}(x)}{1-F_{X}(1)} - frac{F_{X}(1)}{1-F_{X}(1)}$
thus,
$f_{X | X > 1}(x | x > 1) = frac{f_{X}(x)}{1-F_{X}(1)} - 0 = frac{e^{-x}}{e^{-1}} = e^{-(x-1)}$
$E(X | X > 1) = int_{1}^{infty} x f_{X | X > 1} (x | x > 1))dx = int_{1}^{infty} x e^{-(x-1)}dx = 2$ (by int by parts dont really need to show it)
I get how they got the equation $E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
I don't get how they used the CDF to get $f_{X | X > 1}(x | x > 1)$. Any help appreciated.
probability
asked Jan 15 at 23:20
shahshah
1887
$endgroup$
add a comment |
$begingroup$
Could someone please explain the solution:
Solution:
$E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
$F_X(x) = P(X leq x)$
$F_{X | X > 1}(x | x > 1) = P(X leq x | x > 1) = frac{P(X leq x, X > 1)}{P(X > 1)} = frac{F_{X}(x) - F_{X}(1)}{1 - F_{X}(1)}$
$= frac{F_{X}(x)}{1-F_{X}(1)} - frac{F_{X}(1)}{1-F_{X}(1)}$
thus,
$f_{X | X > 1}(x | x > 1) = frac{f_{X}(x)}{1-F_{X}(1)} - 0 = frac{e^{-x}}{e^{-1}} = e^{-(x-1)}$
$E(X | X > 1) = int_{1}^{infty} x f_{X | X > 1} (x | x > 1))dx = int_{1}^{infty} x e^{-(x-1)}dx = 2$ (by int by parts dont really need to show it)
I get how they got the equation $E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
I don't get how they used the CDF to get $f_{X | X > 1}(x | x > 1)$. Any help appreciated.
probability
asked Jan 15 at 23:20
shahshah
1887
$endgroup$
$begingroup$
What is $x$? You have also not specified the distribution governing $X$.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
Sorry fixed it, it was exponential dist with lambda = 1.
$endgroup$
– shah
Jan 15 at 23:53
$begingroup$
PDF is simply derived by differentiating the CDF. The second term in your $F_X(x)$ is constant, so it goes away when you differentiate. The first term is ${F_X(x) over 1-F_X(1)}$, which when differentiated gives ${f_X(x) over 1-F_X(1)}$.
$endgroup$
– Aditya Dua
Jan 16 at 0:29
$begingroup$
ok, but how does $1-F_{X}(1) = e^{-1}$?
$endgroup$
– shah
Jan 16 at 0:39
$begingroup$
Because is $X sim exp(1)$, $F_X(x) = 1-e^{-x}$, so $F_X(1) = 1-e^{-1}$.
$endgroup$
– Aditya Dua
Jan 16 at 6:01
add a comment |
$begingroup$
Could someone please explain the solution:
Solution:
$E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
$F_X(x) = P(X leq x)$
$F_{X | X > 1}(x | x > 1) = P(X leq x | x > 1) = frac{P(X leq x, X > 1)}{P(X > 1)} = frac{F_{X}(x) - F_{X}(1)}{1 - F_{X}(1)}$
$= frac{F_{X}(x)}{1-F_{X}(1)} - frac{F_{X}(1)}{1-F_{X}(1)}$
thus,
$f_{X | X > 1}(x | x > 1) = frac{f_{X}(x)}{1-F_{X}(1)} - 0 = frac{e^{-x}}{e^{-1}} = e^{-(x-1)}$
$E(X | X > 1) = int_{1}^{infty} x f_{X | X > 1} (x | x > 1))dx = int_{1}^{infty} x e^{-(x-1)}dx = 2$ (by int by parts dont really need to show it)
I get how they got the equation $E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
I don't get how they used the CDF to get $f_{X | X > 1}(x | x > 1)$. Any help appreciated.
probability
asked Jan 15 at 23:20
shahshah
1887
$endgroup$
Could someone please explain the solution:
Solution:
$E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
$F_X(x) = P(X leq x)$
$F_{X | X > 1}(x | x > 1) = P(X leq x | x > 1) = frac{P(X leq x, X > 1)}{P(X > 1)} = frac{F_{X}(x) - F_{X}(1)}{1 - F_{X}(1)}$
$= frac{F_{X}(x)}{1-F_{X}(1)} - frac{F_{X}(1)}{1-F_{X}(1)}$
thus,
$f_{X | X > 1}(x | x > 1) = frac{f_{X}(x)}{1-F_{X}(1)} - 0 = frac{e^{-x}}{e^{-1}} = e^{-(x-1)}$
$E(X | X > 1) = int_{1}^{infty} x f_{X | X > 1} (x | x > 1))dx = int_{1}^{infty} x e^{-(x-1)}dx = 2$ (by int by parts dont really need to show it)
I get how they got the equation $E[X | X > 1] = int_{1}^{infty} x f_{X|X>1}(x | x > 1)dx$
I don't get how they used the CDF to get $f_{X | X > 1}(x | x > 1)$. Any help appreciated.
probability
probability
asked Jan 15 at 23:20
shahshah
1887
asked Jan 15 at 23:20
shahshah
1887
edited Jan 15 at 23:53
shah
asked Jan 15 at 23:20
shahshah
1887
asked Jan 15 at 23:20
shahshah
1887
asked Jan 15 at 23:20
shahshah
1887
1887
$begingroup$
What is $x$? You have also not specified the distribution governing $X$.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
Sorry fixed it, it was exponential dist with lambda = 1.
$endgroup$
– shah
Jan 15 at 23:53
$begingroup$
PDF is simply derived by differentiating the CDF. The second term in your $F_X(x)$ is constant, so it goes away when you differentiate. The first term is ${F_X(x) over 1-F_X(1)}$, which when differentiated gives ${f_X(x) over 1-F_X(1)}$.
$endgroup$
– Aditya Dua
Jan 16 at 0:29
$begingroup$
ok, but how does $1-F_{X}(1) = e^{-1}$?
$endgroup$
– shah
Jan 16 at 0:39
$begingroup$
Because is $X sim exp(1)$, $F_X(x) = 1-e^{-x}$, so $F_X(1) = 1-e^{-1}$.
$endgroup$
– Aditya Dua
Jan 16 at 6:01
add a comment |
$begingroup$
What is $x$? You have also not specified the distribution governing $X$.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
Sorry fixed it, it was exponential dist with lambda = 1.
$endgroup$
– shah
Jan 15 at 23:53
$begingroup$
PDF is simply derived by differentiating the CDF. The second term in your $F_X(x)$ is constant, so it goes away when you differentiate. The first term is ${F_X(x) over 1-F_X(1)}$, which when differentiated gives ${f_X(x) over 1-F_X(1)}$.
$endgroup$
– Aditya Dua
Jan 16 at 0:29
$begingroup$
ok, but how does $1-F_{X}(1) = e^{-1}$?
$endgroup$
– shah
Jan 16 at 0:39
$begingroup$
Because is $X sim exp(1)$, $F_X(x) = 1-e^{-x}$, so $F_X(1) = 1-e^{-1}$.
$endgroup$
– Aditya Dua
Jan 16 at 6:01
$begingroup$
What is $x$? You have also not specified the distribution governing $X$.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
What is $x$? You have also not specified the distribution governing $X$.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
Sorry fixed it, it was exponential dist with lambda = 1.
$endgroup$
– shah
Jan 15 at 23:53
$begingroup$
Sorry fixed it, it was exponential dist with lambda = 1.
$endgroup$
– shah
Jan 15 at 23:53
$begingroup$
PDF is simply derived by differentiating the CDF. The second term in your $F_X(x)$ is constant, so it goes away when you differentiate. The first term is ${F_X(x) over 1-F_X(1)}$, which when differentiated gives ${f_X(x) over 1-F_X(1)}$.
$endgroup$
– Aditya Dua
Jan 16 at 0:29
$begingroup$
PDF is simply derived by differentiating the CDF. The second term in your $F_X(x)$ is constant, so it goes away when you differentiate. The first term is ${F_X(x) over 1-F_X(1)}$, which when differentiated gives ${f_X(x) over 1-F_X(1)}$.
$endgroup$
– Aditya Dua
Jan 16 at 0:29
$begingroup$
ok, but how does $1-F_{X}(1) = e^{-1}$?
$endgroup$
– shah
Jan 16 at 0:39
$begingroup$
ok, but how does $1-F_{X}(1) = e^{-1}$?
$endgroup$
– shah
Jan 16 at 0:39
$begingroup$
Because is $X sim exp(1)$, $F_X(x) = 1-e^{-x}$, so $F_X(1) = 1-e^{-1}$.
$endgroup$
– Aditya Dua
Jan 16 at 6:01
$begingroup$
Because is $X sim exp(1)$, $F_X(x) = 1-e^{-x}$, so $F_X(1) = 1-e^{-1}$.
$endgroup$
– Aditya Dua
Jan 16 at 6:01
add a comment |
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$begingroup$
What is $x$? You have also not specified the distribution governing $X$.
$endgroup$
– Aditya Dua
Jan 15 at 23:46
$begingroup$
Sorry fixed it, it was exponential dist with lambda = 1.
$endgroup$
– shah
Jan 15 at 23:53
$begingroup$
PDF is simply derived by differentiating the CDF. The second term in your $F_X(x)$ is constant, so it goes away when you differentiate. The first term is ${F_X(x) over 1-F_X(1)}$, which when differentiated gives ${f_X(x) over 1-F_X(1)}$.
$endgroup$
– Aditya Dua
Jan 16 at 0:29
$begingroup$
ok, but how does $1-F_{X}(1) = e^{-1}$?
$endgroup$
– shah
Jan 16 at 0:39
$begingroup$
Because is $X sim exp(1)$, $F_X(x) = 1-e^{-x}$, so $F_X(1) = 1-e^{-1}$.
$endgroup$
– Aditya Dua
Jan 16 at 6:01