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Why is the expectation of cauchy distribution not defined? (What is the intuition behind it?)
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Let $X$ be random variable with pdf $f_X(x) = dfrac{1}{pi(1+x^2)}$. I understand that mathematically, the improper integral, $displaystyleintlimits_{-infty}^{infty}dfrac{x}{pi(1+x^2)}dx$ does not exist ($underset{T_1to-infty}{lim}underset{T_2toinfty}{lim}displaystyleintlimits_{T_1}^{T_2}dfrac{x}{pi(1+x^2)}dx = frac{ln(1+T_2^2) - ln(1+T_1^2)}{2pi} = infty - infty) implies$ undefined. However I am unable to understand the intuition, why $E[X] ne 0$ since the pdf is an even function which takes positive and negative values with equal probability. Hence large enough samples should produce mean close to 0. Please help?
probability-distributions improper-integrals means
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Let $X$ be random variable with pdf $f_X(x) = dfrac{1}{pi(1+x^2)}$. I understand that mathematically, the improper integral, $displaystyleintlimits_{-infty}^{infty}dfrac{x}{pi(1+x^2)}dx$ does not exist ($underset{T_1to-infty}{lim}underset{T_2toinfty}{lim}displaystyleintlimits_{T_1}^{T_2}dfrac{x}{pi(1+x^2)}dx = frac{ln(1+T_2^2) - ln(1+T_1^2)}{2pi} = infty - infty) implies$ undefined. However I am unable to understand the intuition, why $E[X] ne 0$ since the pdf is an even function which takes positive and negative values with equal probability. Hence large enough samples should produce mean close to 0. Please help?
probability-distributions improper-integrals means
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sh10
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2
According to your logic, large enough samples should agree with the central limit theorem, too. But that is not the case for the Cauchy distribution: $frac{x}{x^2+1}notin L^1(mathbb{R})$, full stop.
– Jack D'Aurizio
yesterday
It is well-known fact, that for Cauchy distribution, arithmetic mean of independent samples $frac{X_1+ldots+X_n}{n}$ is also Cauchy distributed with the same pdf as of summands. So, the mean of large enough samples is not close to zero. It behaves as a single sample from this distribution.
– NCh
yesterday
add a comment |
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Let $X$ be random variable with pdf $f_X(x) = dfrac{1}{pi(1+x^2)}$. I understand that mathematically, the improper integral, $displaystyleintlimits_{-infty}^{infty}dfrac{x}{pi(1+x^2)}dx$ does not exist ($underset{T_1to-infty}{lim}underset{T_2toinfty}{lim}displaystyleintlimits_{T_1}^{T_2}dfrac{x}{pi(1+x^2)}dx = frac{ln(1+T_2^2) - ln(1+T_1^2)}{2pi} = infty - infty) implies$ undefined. However I am unable to understand the intuition, why $E[X] ne 0$ since the pdf is an even function which takes positive and negative values with equal probability. Hence large enough samples should produce mean close to 0. Please help?
probability-distributions improper-integrals means
asked yesterday
sh10
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sh10 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $X$ be random variable with pdf $f_X(x) = dfrac{1}{pi(1+x^2)}$. I understand that mathematically, the improper integral, $displaystyleintlimits_{-infty}^{infty}dfrac{x}{pi(1+x^2)}dx$ does not exist ($underset{T_1to-infty}{lim}underset{T_2toinfty}{lim}displaystyleintlimits_{T_1}^{T_2}dfrac{x}{pi(1+x^2)}dx = frac{ln(1+T_2^2) - ln(1+T_1^2)}{2pi} = infty - infty) implies$ undefined. However I am unable to understand the intuition, why $E[X] ne 0$ since the pdf is an even function which takes positive and negative values with equal probability. Hence large enough samples should produce mean close to 0. Please help?
probability-distributions improper-integrals means
probability-distributions improper-integrals means
asked yesterday
sh10
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asked yesterday
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edited 21 hours ago
asked yesterday
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asked yesterday
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asked yesterday
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2
According to your logic, large enough samples should agree with the central limit theorem, too. But that is not the case for the Cauchy distribution: $frac{x}{x^2+1}notin L^1(mathbb{R})$, full stop.
– Jack D'Aurizio
yesterday
It is well-known fact, that for Cauchy distribution, arithmetic mean of independent samples $frac{X_1+ldots+X_n}{n}$ is also Cauchy distributed with the same pdf as of summands. So, the mean of large enough samples is not close to zero. It behaves as a single sample from this distribution.
– NCh
yesterday
add a comment |
2
According to your logic, large enough samples should agree with the central limit theorem, too. But that is not the case for the Cauchy distribution: $frac{x}{x^2+1}notin L^1(mathbb{R})$, full stop.
– Jack D'Aurizio
yesterday
It is well-known fact, that for Cauchy distribution, arithmetic mean of independent samples $frac{X_1+ldots+X_n}{n}$ is also Cauchy distributed with the same pdf as of summands. So, the mean of large enough samples is not close to zero. It behaves as a single sample from this distribution.
– NCh
yesterday
2
2
According to your logic, large enough samples should agree with the central limit theorem, too. But that is not the case for the Cauchy distribution: $frac{x}{x^2+1}notin L^1(mathbb{R})$, full stop.
– Jack D'Aurizio
yesterday
According to your logic, large enough samples should agree with the central limit theorem, too. But that is not the case for the Cauchy distribution: $frac{x}{x^2+1}notin L^1(mathbb{R})$, full stop.
– Jack D'Aurizio
yesterday
It is well-known fact, that for Cauchy distribution, arithmetic mean of independent samples $frac{X_1+ldots+X_n}{n}$ is also Cauchy distributed with the same pdf as of summands. So, the mean of large enough samples is not close to zero. It behaves as a single sample from this distribution.
– NCh
yesterday
It is well-known fact, that for Cauchy distribution, arithmetic mean of independent samples $frac{X_1+ldots+X_n}{n}$ is also Cauchy distributed with the same pdf as of summands. So, the mean of large enough samples is not close to zero. It behaves as a single sample from this distribution.
– NCh
yesterday
add a comment |
1 Answer
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There are several ways to look at it:
- Let $f(x):=frac{pi^{-1}}{1+x^2}$ so $int_{-infty}^c xf(x) dx=-infty,,int_d^infty xf(x) dx$ for any $c,,dinmathbb{R}$. So if we choose $c<d$, you could argue the mean is $-infty+int_c^d xf(x) dx+infty$. In theory, you can get any value you like if you think the infinities cancel.
- "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $infty-infty$ is an indefinite form, so you can't use that theorem.
- The characteristic function is $varphi(t):=exp -left|tright|$. If we average $n$ samples, the result has characteristic function $varphi^n(t/n)=varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $mu$, no $(X-mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $mu=varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $lim_{tto 0^pm}frac{|t|}{t}$ differ).
- It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.
answered 21 hours ago
J.G.
18.2k21932
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1 Answer
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up vote
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There are several ways to look at it:
- Let $f(x):=frac{pi^{-1}}{1+x^2}$ so $int_{-infty}^c xf(x) dx=-infty,,int_d^infty xf(x) dx$ for any $c,,dinmathbb{R}$. So if we choose $c<d$, you could argue the mean is $-infty+int_c^d xf(x) dx+infty$. In theory, you can get any value you like if you think the infinities cancel.
- "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $infty-infty$ is an indefinite form, so you can't use that theorem.
- The characteristic function is $varphi(t):=exp -left|tright|$. If we average $n$ samples, the result has characteristic function $varphi^n(t/n)=varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $mu$, no $(X-mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $mu=varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $lim_{tto 0^pm}frac{|t|}{t}$ differ).
- It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.
answered 21 hours ago
J.G.
18.2k21932
add a comment |
up vote
1
down vote
There are several ways to look at it:
- Let $f(x):=frac{pi^{-1}}{1+x^2}$ so $int_{-infty}^c xf(x) dx=-infty,,int_d^infty xf(x) dx$ for any $c,,dinmathbb{R}$. So if we choose $c<d$, you could argue the mean is $-infty+int_c^d xf(x) dx+infty$. In theory, you can get any value you like if you think the infinities cancel.
- "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $infty-infty$ is an indefinite form, so you can't use that theorem.
- The characteristic function is $varphi(t):=exp -left|tright|$. If we average $n$ samples, the result has characteristic function $varphi^n(t/n)=varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $mu$, no $(X-mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $mu=varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $lim_{tto 0^pm}frac{|t|}{t}$ differ).
- It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.
answered 21 hours ago
J.G.
18.2k21932
add a comment |
up vote
1
down vote
up vote
1
down vote
There are several ways to look at it:
- Let $f(x):=frac{pi^{-1}}{1+x^2}$ so $int_{-infty}^c xf(x) dx=-infty,,int_d^infty xf(x) dx$ for any $c,,dinmathbb{R}$. So if we choose $c<d$, you could argue the mean is $-infty+int_c^d xf(x) dx+infty$. In theory, you can get any value you like if you think the infinities cancel.
- "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $infty-infty$ is an indefinite form, so you can't use that theorem.
- The characteristic function is $varphi(t):=exp -left|tright|$. If we average $n$ samples, the result has characteristic function $varphi^n(t/n)=varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $mu$, no $(X-mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $mu=varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $lim_{tto 0^pm}frac{|t|}{t}$ differ).
- It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.
answered 21 hours ago
J.G.
18.2k21932
There are several ways to look at it:
- Let $f(x):=frac{pi^{-1}}{1+x^2}$ so $int_{-infty}^c xf(x) dx=-infty,,int_d^infty xf(x) dx$ for any $c,,dinmathbb{R}$. So if we choose $c<d$, you could argue the mean is $-infty+int_c^d xf(x) dx+infty$. In theory, you can get any value you like if you think the infinities cancel.
- "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $infty-infty$ is an indefinite form, so you can't use that theorem.
- The characteristic function is $varphi(t):=exp -left|tright|$. If we average $n$ samples, the result has characteristic function $varphi^n(t/n)=varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $mu$, no $(X-mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $mu=varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $lim_{tto 0^pm}frac{|t|}{t}$ differ).
- It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.
answered 21 hours ago
J.G.
18.2k21932
answered 21 hours ago
J.G.
18.2k21932
answered 21 hours ago
J.G.
18.2k21932
answered 21 hours ago
J.G.
18.2k21932
18.2k21932
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2
According to your logic, large enough samples should agree with the central limit theorem, too. But that is not the case for the Cauchy distribution: $frac{x}{x^2+1}notin L^1(mathbb{R})$, full stop.
– Jack D'Aurizio
yesterday
It is well-known fact, that for Cauchy distribution, arithmetic mean of independent samples $frac{X_1+ldots+X_n}{n}$ is also Cauchy distributed with the same pdf as of summands. So, the mean of large enough samples is not close to zero. It behaves as a single sample from this distribution.
– NCh
yesterday