Limits of a density functionWriting the density of a continuous random variable in terms of a...
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Limits of a density function
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$begingroup$
If the limit of a density function exists does it the follow that it is zero? To put is formally
$$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$
asked 4 hours ago
Jesper HybelJesper Hybel
921614
$endgroup$
add a comment |
$begingroup$
If the limit of a density function exists does it the follow that it is zero? To put is formally
$$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$
asked 4 hours ago
Jesper HybelJesper Hybel
921614
$endgroup$
add a comment |
$begingroup$
If the limit of a density function exists does it the follow that it is zero? To put is formally
$$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$
asked 4 hours ago
Jesper HybelJesper Hybel
921614
$endgroup$
If the limit of a density function exists does it the follow that it is zero? To put is formally
$$exists a in mathbb R lim_{t rightarrow infty} f(t) = a Rightarrow a= 0.$$
asked 4 hours ago
Jesper HybelJesper Hybel
921614
asked 4 hours ago
Jesper HybelJesper Hybel
921614
asked 4 hours ago
Jesper HybelJesper Hybel
921614
asked 4 hours ago
Jesper HybelJesper Hybel
921614
asked 4 hours ago
Jesper HybelJesper Hybel
921614
921614
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Yes.
Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.
But then:
$$
int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
$$
So $f$ cannot be a density function.
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes.
Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.
But then:
$$
int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
$$
So $f$ cannot be a density function.
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
$endgroup$
add a comment |
$begingroup$
Yes.
Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.
But then:
$$
int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
$$
So $f$ cannot be a density function.
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
$endgroup$
add a comment |
$begingroup$
Yes.
Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.
But then:
$$
int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
$$
So $f$ cannot be a density function.
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
$endgroup$
Yes.
Suppose the limit is anything else, so $lim_{t rightarrow infty} f(t) = a neq 0$. Then, by the definition of the limit, there is an $N$ so that for all $t > N$, $| f(t) - a | < frac{a}{2}$. In particular, $f(t) > frac{a}{2}$ in this reigon.
But then:
$$
int_{mathbf{R}} f(t) dt geq int_{N}^{infty} f(t) dt geq int_{N}^{infty} frac{a}{2} dt = infty
$$
So $f$ cannot be a density function.
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
edited 2 hours ago
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
answered 3 hours ago
Matthew DruryMatthew Drury
25.8k262104
25.8k262104
add a comment |
add a comment |
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