Moment-generating function of a random variable with mean 0 and variance 1
Consider a random variable $X$ with mean $0$ and variance $1$. Given $0 < d < 1$, do we have
$$E(e^{d X}) leq 1 + O(d^2)$$
(it's true when $|X| leq 1$)
probability probability-distributions moment-generating-functions
asked Nov 6 at 16:09
permanganate
1537
add a comment |
Consider a random variable $X$ with mean $0$ and variance $1$. Given $0 < d < 1$, do we have
$$E(e^{d X}) leq 1 + O(d^2)$$
(it's true when $|X| leq 1$)
probability probability-distributions moment-generating-functions
asked Nov 6 at 16:09
permanganate
1537
No, $mathbb{E}(exp(dX))$ need not exist for $d>0$.
– user10354138
Nov 6 at 16:13
add a comment |
Consider a random variable $X$ with mean $0$ and variance $1$. Given $0 < d < 1$, do we have
$$E(e^{d X}) leq 1 + O(d^2)$$
(it's true when $|X| leq 1$)
probability probability-distributions moment-generating-functions
asked Nov 6 at 16:09
permanganate
1537
Consider a random variable $X$ with mean $0$ and variance $1$. Given $0 < d < 1$, do we have
$$E(e^{d X}) leq 1 + O(d^2)$$
(it's true when $|X| leq 1$)
probability probability-distributions moment-generating-functions
probability probability-distributions moment-generating-functions
asked Nov 6 at 16:09
permanganate
1537
asked Nov 6 at 16:09
permanganate
1537
asked Nov 6 at 16:09
permanganate
1537
asked Nov 6 at 16:09
permanganate
1537
asked Nov 6 at 16:09
permanganate
1537
1537
No, $mathbb{E}(exp(dX))$ need not exist for $d>0$.
– user10354138
Nov 6 at 16:13
add a comment |
No, $mathbb{E}(exp(dX))$ need not exist for $d>0$.
– user10354138
Nov 6 at 16:13
No, $mathbb{E}(exp(dX))$ need not exist for $d>0$.
– user10354138
Nov 6 at 16:13
No, $mathbb{E}(exp(dX))$ need not exist for $d>0$.
– user10354138
Nov 6 at 16:13
add a comment |
1 Answer
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Consider $X sim N(0,1)$.
We know that $mathbb{E}[e^{dX}]=e^{d^2/2}$, which is clearly not $O(d^2)$. So the answer is no, based on this counter-example.
answered Nov 23 at 5:38
Aditya Dua
80418
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
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
Consider $X sim N(0,1)$.
We know that $mathbb{E}[e^{dX}]=e^{d^2/2}$, which is clearly not $O(d^2)$. So the answer is no, based on this counter-example.
answered Nov 23 at 5:38
Aditya Dua
80418
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
add a comment |
Consider $X sim N(0,1)$.
We know that $mathbb{E}[e^{dX}]=e^{d^2/2}$, which is clearly not $O(d^2)$. So the answer is no, based on this counter-example.
answered Nov 23 at 5:38
Aditya Dua
80418
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
add a comment |
Consider $X sim N(0,1)$.
We know that $mathbb{E}[e^{dX}]=e^{d^2/2}$, which is clearly not $O(d^2)$. So the answer is no, based on this counter-example.
answered Nov 23 at 5:38
Aditya Dua
80418
Consider $X sim N(0,1)$.
We know that $mathbb{E}[e^{dX}]=e^{d^2/2}$, which is clearly not $O(d^2)$. So the answer is no, based on this counter-example.
answered Nov 23 at 5:38
Aditya Dua
80418
answered Nov 23 at 5:38
Aditya Dua
80418
answered Nov 23 at 5:38
Aditya Dua
80418
answered Nov 23 at 5:38
Aditya Dua
80418
80418
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
add a comment |
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
It is $e^{d^2/2} leq 1 + d^2$ when $0 < d < 1$ (I didn't ask for $E(e^{dX})$ to be $O(d^2)$ but $1+O(d^2)$).
– permanganate
Nov 23 at 10:30
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
Interesting, I wonder if the normal RV example can be used to construct a proof...
– Aditya Dua
Nov 23 at 23:42
add a comment |
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No, $mathbb{E}(exp(dX))$ need not exist for $d>0$.
– user10354138
Nov 6 at 16:13