Mixing
Calculating $mathbb{E}[Re^{iX}]$ where $R$ and $X$ are independent random variables.
$begingroup$
Let $R$ be an exponential random variable with parameter $lambda>0$ and let X be independent from R and uniformly distributed on $(0,2pi)$, find $mathbb{E}[Re^{iX}]$. My intuition tells me that the expectation of the product is the product of the expectations and that that the final answer should be $-mathbb{E}[R]$ since $mathbb{E}[X]=pi$. But using Euler's formula and integrating I get that $mathbb{E}[e^{iX}]=0$ which doesn't make sense. Any tips on this would be appreciated.
probability integration probability-theory complex-numbers expected-value
asked Jan 27 at 13:36
FedericoFederico
83
$endgroup$
add a comment |
$begingroup$
Let $R$ be an exponential random variable with parameter $lambda>0$ and let X be independent from R and uniformly distributed on $(0,2pi)$, find $mathbb{E}[Re^{iX}]$. My intuition tells me that the expectation of the product is the product of the expectations and that that the final answer should be $-mathbb{E}[R]$ since $mathbb{E}[X]=pi$. But using Euler's formula and integrating I get that $mathbb{E}[e^{iX}]=0$ which doesn't make sense. Any tips on this would be appreciated.
probability integration probability-theory complex-numbers expected-value
asked Jan 27 at 13:36
FedericoFederico
83
$endgroup$
add a comment |
$begingroup$
Let $R$ be an exponential random variable with parameter $lambda>0$ and let X be independent from R and uniformly distributed on $(0,2pi)$, find $mathbb{E}[Re^{iX}]$. My intuition tells me that the expectation of the product is the product of the expectations and that that the final answer should be $-mathbb{E}[R]$ since $mathbb{E}[X]=pi$. But using Euler's formula and integrating I get that $mathbb{E}[e^{iX}]=0$ which doesn't make sense. Any tips on this would be appreciated.
probability integration probability-theory complex-numbers expected-value
asked Jan 27 at 13:36
FedericoFederico
83
$endgroup$
Let $R$ be an exponential random variable with parameter $lambda>0$ and let X be independent from R and uniformly distributed on $(0,2pi)$, find $mathbb{E}[Re^{iX}]$. My intuition tells me that the expectation of the product is the product of the expectations and that that the final answer should be $-mathbb{E}[R]$ since $mathbb{E}[X]=pi$. But using Euler's formula and integrating I get that $mathbb{E}[e^{iX}]=0$ which doesn't make sense. Any tips on this would be appreciated.
probability integration probability-theory complex-numbers expected-value
probability integration probability-theory complex-numbers expected-value
asked Jan 27 at 13:36
FedericoFederico
83
asked Jan 27 at 13:36
FedericoFederico
83
edited Jan 27 at 13:49
Federico
asked Jan 27 at 13:36
FedericoFederico
83
asked Jan 27 at 13:36
FedericoFederico
83
asked Jan 27 at 13:36
FedericoFederico
83
83
add a comment |
add a comment |
1 Answer
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$begingroup$
Your wrong assumption seems to be that $mathbb Ee^{iX}=e^{mathbb EiX}=e^{ipi}=-1$.
The first equality is not correct.
It is correct that $mathbb Ee^{iX}=0$.
Observe that $e^{iX}$ and $-e^{iX}=e^{i(X+pi)}$ will have equal distribution (both uniform on unit circle).
This tells us that $-mathbb Ee^{iX}=mathbb Ee^{iX}$ or equivalently $mathbb Ee^{iX}=0$.
answered Jan 27 at 13:49
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your wrong assumption seems to be that $mathbb Ee^{iX}=e^{mathbb EiX}=e^{ipi}=-1$.
The first equality is not correct.
It is correct that $mathbb Ee^{iX}=0$.
Observe that $e^{iX}$ and $-e^{iX}=e^{i(X+pi)}$ will have equal distribution (both uniform on unit circle).
This tells us that $-mathbb Ee^{iX}=mathbb Ee^{iX}$ or equivalently $mathbb Ee^{iX}=0$.
answered Jan 27 at 13:49
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
add a comment |
$begingroup$
Your wrong assumption seems to be that $mathbb Ee^{iX}=e^{mathbb EiX}=e^{ipi}=-1$.
The first equality is not correct.
It is correct that $mathbb Ee^{iX}=0$.
Observe that $e^{iX}$ and $-e^{iX}=e^{i(X+pi)}$ will have equal distribution (both uniform on unit circle).
This tells us that $-mathbb Ee^{iX}=mathbb Ee^{iX}$ or equivalently $mathbb Ee^{iX}=0$.
answered Jan 27 at 13:49
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
add a comment |
$begingroup$
Your wrong assumption seems to be that $mathbb Ee^{iX}=e^{mathbb EiX}=e^{ipi}=-1$.
The first equality is not correct.
It is correct that $mathbb Ee^{iX}=0$.
Observe that $e^{iX}$ and $-e^{iX}=e^{i(X+pi)}$ will have equal distribution (both uniform on unit circle).
This tells us that $-mathbb Ee^{iX}=mathbb Ee^{iX}$ or equivalently $mathbb Ee^{iX}=0$.
answered Jan 27 at 13:49
drhabdrhab
103k545136
$endgroup$
Your wrong assumption seems to be that $mathbb Ee^{iX}=e^{mathbb EiX}=e^{ipi}=-1$.
The first equality is not correct.
It is correct that $mathbb Ee^{iX}=0$.
Observe that $e^{iX}$ and $-e^{iX}=e^{i(X+pi)}$ will have equal distribution (both uniform on unit circle).
This tells us that $-mathbb Ee^{iX}=mathbb Ee^{iX}$ or equivalently $mathbb Ee^{iX}=0$.
answered Jan 27 at 13:49
drhabdrhab
103k545136
answered Jan 27 at 13:49
drhabdrhab
103k545136
answered Jan 27 at 13:49
drhabdrhab
103k545136
answered Jan 27 at 13:49
drhabdrhab
103k545136
103k545136
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
add a comment |
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Thanks for the reply, also if we let $Z=Re^{iX}$ and I have to calculate, for example, $mathbb{E}[|Z|^2]$, is it justifiable to say $mathbb{E}[|Z|^2]=mathbb{E}[|Re^{iX}|^2]=mathbb{E}[|R|^2]$?
$endgroup$
– Federico
Jan 27 at 14:37
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
$begingroup$
Yes, that is correct. And moreover $|R|^2=R^2$ so also $cdots=mathbb ER^2$.
$endgroup$
– drhab
Jan 27 at 14:40
add a comment |
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