Conditional expectation of exponential rv
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Let $ X $~$Expo(lambda)$. Find $E(X|X < 1) $.
I can find the conditional diaribution of$X|X<1$ and then evaluate the sum. But the question wants me to find the solution using law of total expectation and i don't know how to do it without using calculus. Any help would be appreciated!
probability probability-theory
asked Nov 24 at 5:28
user587126
155
add a comment |
up vote
0
down vote
favorite
Let $ X $~$Expo(lambda)$. Find $E(X|X < 1) $.
I can find the conditional diaribution of$X|X<1$ and then evaluate the sum. But the question wants me to find the solution using law of total expectation and i don't know how to do it without using calculus. Any help would be appreciated!
probability probability-theory
asked Nov 24 at 5:28
user587126
155
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $ X $~$Expo(lambda)$. Find $E(X|X < 1) $.
I can find the conditional diaribution of$X|X<1$ and then evaluate the sum. But the question wants me to find the solution using law of total expectation and i don't know how to do it without using calculus. Any help would be appreciated!
probability probability-theory
asked Nov 24 at 5:28
user587126
155
Let $ X $~$Expo(lambda)$. Find $E(X|X < 1) $.
I can find the conditional diaribution of$X|X<1$ and then evaluate the sum. But the question wants me to find the solution using law of total expectation and i don't know how to do it without using calculus. Any help would be appreciated!
probability probability-theory
probability probability-theory
asked Nov 24 at 5:28
user587126
155
asked Nov 24 at 5:28
user587126
155
asked Nov 24 at 5:28
user587126
155
asked Nov 24 at 5:28
user587126
155
asked Nov 24 at 5:28
user587126
155
155
add a comment |
add a comment |
1 Answer
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0
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accepted
begin{align}E[X]&=E[X|X>1]P(X>1)+E[X|X<1]P(X<1) \
&=(1+E[X])P(X>1)+E[X|X<1]P(X<1)end{align}
Where I have used the memoryless property.
Hopefully you can take it from here.
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
begin{align}E[X]&=E[X|X>1]P(X>1)+E[X|X<1]P(X<1) \
&=(1+E[X])P(X>1)+E[X|X<1]P(X<1)end{align}
Where I have used the memoryless property.
Hopefully you can take it from here.
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
|
show 1 more comment
up vote
0
down vote
accepted
begin{align}E[X]&=E[X|X>1]P(X>1)+E[X|X<1]P(X<1) \
&=(1+E[X])P(X>1)+E[X|X<1]P(X<1)end{align}
Where I have used the memoryless property.
Hopefully you can take it from here.
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
|
show 1 more comment
up vote
0
down vote
accepted
up vote
0
down vote
accepted
begin{align}E[X]&=E[X|X>1]P(X>1)+E[X|X<1]P(X<1) \
&=(1+E[X])P(X>1)+E[X|X<1]P(X<1)end{align}
Where I have used the memoryless property.
Hopefully you can take it from here.
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
begin{align}E[X]&=E[X|X>1]P(X>1)+E[X|X<1]P(X<1) \
&=(1+E[X])P(X>1)+E[X|X<1]P(X<1)end{align}
Where I have used the memoryless property.
Hopefully you can take it from here.
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
answered Nov 24 at 5:35
Siong Thye Goh
98.2k1463116
98.2k1463116
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
|
show 1 more comment
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Thanks! It is really difficult to remember all the properties of a certain distribution while solving problems!
– user587126
Nov 24 at 6:53
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
Can you please elaborate how you used the memoryless property?
– user587126
Nov 24 at 7:09
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
memoryless property tells us $E[X|X>x] = x + E[X]$.
– Siong Thye Goh
Nov 24 at 7:25
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
See stats.stackexchange.com/questions/48496/… for a proof.
– N74
Nov 24 at 7:27
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
But how do you get this from $P(X>s+t|X>t)=P(X>s)$?
– user587126
Nov 24 at 7:29
|
show 1 more comment
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