Calculate the expectation of random variable, integration of a function.
$begingroup$
I want to calculate the expectation of one random variable $frac{1}{sqrt{a^2+x^2}}$, where $xsim N(0,sigma^2)$ and $a$ is a constant.
It is straightforward that we can come to the integration of the following
$$intlimits _{-infty}^{infty} exp(-frac{x^2}{2sigma^2}) frac{1}{sqrt{x^2+a^2}} dx.$$
How can we handle this integration, or how can we approximate its value?
probability integration probability-distributions expected-value
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
$endgroup$
add a comment |
$begingroup$
I want to calculate the expectation of one random variable $frac{1}{sqrt{a^2+x^2}}$, where $xsim N(0,sigma^2)$ and $a$ is a constant.
It is straightforward that we can come to the integration of the following
$$intlimits _{-infty}^{infty} exp(-frac{x^2}{2sigma^2}) frac{1}{sqrt{x^2+a^2}} dx.$$
How can we handle this integration, or how can we approximate its value?
probability integration probability-distributions expected-value
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
$endgroup$
$begingroup$
This is a modified Bessel function in terms of the quantity $eta = a^2/(2sigma)^2$, which has a nice series expansion about $eta = infty$, so it would have good approximation properties when $a/sigma$ is large.
$endgroup$
– heropup
Dec 11 '18 at 8:44
$begingroup$
Thanks! What is the general form of the modified Bessel function you mentioned?
$endgroup$
– ZHANG Wei
Dec 11 '18 at 9:00
add a comment |
$begingroup$
I want to calculate the expectation of one random variable $frac{1}{sqrt{a^2+x^2}}$, where $xsim N(0,sigma^2)$ and $a$ is a constant.
It is straightforward that we can come to the integration of the following
$$intlimits _{-infty}^{infty} exp(-frac{x^2}{2sigma^2}) frac{1}{sqrt{x^2+a^2}} dx.$$
How can we handle this integration, or how can we approximate its value?
probability integration probability-distributions expected-value
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
$endgroup$
I want to calculate the expectation of one random variable $frac{1}{sqrt{a^2+x^2}}$, where $xsim N(0,sigma^2)$ and $a$ is a constant.
It is straightforward that we can come to the integration of the following
$$intlimits _{-infty}^{infty} exp(-frac{x^2}{2sigma^2}) frac{1}{sqrt{x^2+a^2}} dx.$$
How can we handle this integration, or how can we approximate its value?
probability integration probability-distributions expected-value
probability integration probability-distributions expected-value
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
asked Dec 11 '18 at 8:21
ZHANG WeiZHANG Wei
388
388
$begingroup$
This is a modified Bessel function in terms of the quantity $eta = a^2/(2sigma)^2$, which has a nice series expansion about $eta = infty$, so it would have good approximation properties when $a/sigma$ is large.
$endgroup$
– heropup
Dec 11 '18 at 8:44
$begingroup$
Thanks! What is the general form of the modified Bessel function you mentioned?
$endgroup$
– ZHANG Wei
Dec 11 '18 at 9:00
add a comment |
$begingroup$
This is a modified Bessel function in terms of the quantity $eta = a^2/(2sigma)^2$, which has a nice series expansion about $eta = infty$, so it would have good approximation properties when $a/sigma$ is large.
$endgroup$
– heropup
Dec 11 '18 at 8:44
$begingroup$
Thanks! What is the general form of the modified Bessel function you mentioned?
$endgroup$
– ZHANG Wei
Dec 11 '18 at 9:00
$begingroup$
This is a modified Bessel function in terms of the quantity $eta = a^2/(2sigma)^2$, which has a nice series expansion about $eta = infty$, so it would have good approximation properties when $a/sigma$ is large.
$endgroup$
– heropup
Dec 11 '18 at 8:44
$begingroup$
This is a modified Bessel function in terms of the quantity $eta = a^2/(2sigma)^2$, which has a nice series expansion about $eta = infty$, so it would have good approximation properties when $a/sigma$ is large.
$endgroup$
– heropup
Dec 11 '18 at 8:44
$begingroup$
Thanks! What is the general form of the modified Bessel function you mentioned?
$endgroup$
– ZHANG Wei
Dec 11 '18 at 9:00
$begingroup$
Thanks! What is the general form of the modified Bessel function you mentioned?
$endgroup$
– ZHANG Wei
Dec 11 '18 at 9:00
add a comment |
0
active
oldest
votes
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%2f3035054%2fcalculate-the-expectation-of-random-variable-integration-of-a-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3035054%2fcalculate-the-expectation-of-random-variable-integration-of-a-function%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
$begingroup$
This is a modified Bessel function in terms of the quantity $eta = a^2/(2sigma)^2$, which has a nice series expansion about $eta = infty$, so it would have good approximation properties when $a/sigma$ is large.
$endgroup$
– heropup
Dec 11 '18 at 8:44
$begingroup$
Thanks! What is the general form of the modified Bessel function you mentioned?
$endgroup$
– ZHANG Wei
Dec 11 '18 at 9:00