Mixing
Distribution of X SQRT(XX+YY) where X and Y are Gaussian
$begingroup$
I have a complex variable $Z = X + i Y$ where $X$ and $Y$ are Gaussian iid with zero mean and $sigma^2$ variance.
I am interested in ${mathbb E} ( X |Z|^p )$.
Is there a known distribution for this?
I know that $|Z|$ is Rayleigh distributed. However, for my expression I would have ${mathbb E} (X+iY)|Z|^p = {mathbb E} X|Z|^p + i {mathbb E} Y|Z|^p = (1+i) {mathbb E} X|Z|^p$.
probability probability-distributions
asked Jan 8 at 18:43
divBdivB
228111
$endgroup$
add a comment |
$begingroup$
I have a complex variable $Z = X + i Y$ where $X$ and $Y$ are Gaussian iid with zero mean and $sigma^2$ variance.
I am interested in ${mathbb E} ( X |Z|^p )$.
Is there a known distribution for this?
I know that $|Z|$ is Rayleigh distributed. However, for my expression I would have ${mathbb E} (X+iY)|Z|^p = {mathbb E} X|Z|^p + i {mathbb E} Y|Z|^p = (1+i) {mathbb E} X|Z|^p$.
probability probability-distributions
asked Jan 8 at 18:43
divBdivB
228111
$endgroup$
add a comment |
$begingroup$
I have a complex variable $Z = X + i Y$ where $X$ and $Y$ are Gaussian iid with zero mean and $sigma^2$ variance.
I am interested in ${mathbb E} ( X |Z|^p )$.
Is there a known distribution for this?
I know that $|Z|$ is Rayleigh distributed. However, for my expression I would have ${mathbb E} (X+iY)|Z|^p = {mathbb E} X|Z|^p + i {mathbb E} Y|Z|^p = (1+i) {mathbb E} X|Z|^p$.
probability probability-distributions
asked Jan 8 at 18:43
divBdivB
228111
$endgroup$
I have a complex variable $Z = X + i Y$ where $X$ and $Y$ are Gaussian iid with zero mean and $sigma^2$ variance.
I am interested in ${mathbb E} ( X |Z|^p )$.
Is there a known distribution for this?
I know that $|Z|$ is Rayleigh distributed. However, for my expression I would have ${mathbb E} (X+iY)|Z|^p = {mathbb E} X|Z|^p + i {mathbb E} Y|Z|^p = (1+i) {mathbb E} X|Z|^p$.
probability probability-distributions
probability probability-distributions
asked Jan 8 at 18:43
divBdivB
228111
asked Jan 8 at 18:43
divBdivB
228111
edited Jan 8 at 18:52
divB
asked Jan 8 at 18:43
divBdivB
228111
asked Jan 8 at 18:43
divBdivB
228111
asked Jan 8 at 18:43
divBdivB
228111
228111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write $Z=Rexp iTheta$, where in requiring $Thetain [0,,2pi]$ we impose a uniform distribution on $Theta$. You're studying $R^{p+1}cosTheta$, a product of two independent variables. In particular, the mean is $0$ because the cosine averages to $0$.
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
$endgroup$
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
Hmm I think I found the trouble somewhere else ...x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x)gives me zero on average.
$endgroup$
– divB
Jan 8 at 22:48
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066566%2fdistribution-of-x-sqrtxxyy-where-x-and-y-are-gaussian%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write $Z=Rexp iTheta$, where in requiring $Thetain [0,,2pi]$ we impose a uniform distribution on $Theta$. You're studying $R^{p+1}cosTheta$, a product of two independent variables. In particular, the mean is $0$ because the cosine averages to $0$.
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
$endgroup$
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
Hmm I think I found the trouble somewhere else ...x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x)gives me zero on average.
$endgroup$
– divB
Jan 8 at 22:48
add a comment |
$begingroup$
Write $Z=Rexp iTheta$, where in requiring $Thetain [0,,2pi]$ we impose a uniform distribution on $Theta$. You're studying $R^{p+1}cosTheta$, a product of two independent variables. In particular, the mean is $0$ because the cosine averages to $0$.
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
$endgroup$
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
Hmm I think I found the trouble somewhere else ...x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x)gives me zero on average.
$endgroup$
– divB
Jan 8 at 22:48
add a comment |
$begingroup$
Write $Z=Rexp iTheta$, where in requiring $Thetain [0,,2pi]$ we impose a uniform distribution on $Theta$. You're studying $R^{p+1}cosTheta$, a product of two independent variables. In particular, the mean is $0$ because the cosine averages to $0$.
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
$endgroup$
Write $Z=Rexp iTheta$, where in requiring $Thetain [0,,2pi]$ we impose a uniform distribution on $Theta$. You're studying $R^{p+1}cosTheta$, a product of two independent variables. In particular, the mean is $0$ because the cosine averages to $0$.
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
answered Jan 8 at 18:56
J.G.J.G.
25.3k22539
25.3k22539
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
Hmm I think I found the trouble somewhere else ...x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x)gives me zero on average.
$endgroup$
– divB
Jan 8 at 22:48
add a comment |
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
Hmm I think I found the trouble somewhere else ...x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x)gives me zero on average.
$endgroup$
– divB
Jan 8 at 22:48
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
That's a great input! Actually I want ${mathbb E}(|Z|^p Z)$ anway (also ${mathbb E}((|Z|^p Z)^2)$). So with $Z=|Z|e^{i operatorname{arg}Z}$ I am writing ${mathbb E}(R e^{iphi})$ = ${mathbb E}(R) {mathbb E}(e^{iphi})$ with $R$ Rayleigh and $phi$ uniformly distributed (and independent). Now I am trying to obtain ${mathbb E}((e^{iphi})^q) = frac{1}{2pi} int_{-pi}^{pi} e^{i q phi} dphi = operatorname{sinc}(q)$ which is always zero. What am I missing?
$endgroup$
– divB
Jan 8 at 22:21
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
@divB Look beyond integer-valued $q$ if you want something non-trivial.
$endgroup$
– J.G.
Jan 8 at 22:25
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
I know sinc is nonzero for non-integers but $q$ just happens to be 1 or 2 ... and zero doesn't make sense nor does it match with the numerics in MATLAB. I assume the assumption of independence is not true or so?
$endgroup$
– divB
Jan 8 at 22:35
$begingroup$
Hmm I think I found the trouble somewhere else ...
x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x) gives me zero on average.$endgroup$
– divB
Jan 8 at 22:48
$begingroup$
Hmm I think I found the trouble somewhere else ...
x=sigma*(randn(N,1)+i*randn(N,1)); mean(abs(x)^p .* x) gives me zero on average.$endgroup$
– divB
Jan 8 at 22:48
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066566%2fdistribution-of-x-sqrtxxyy-where-x-and-y-are-gaussian%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
