Limitation on the difference of characteristic functions
Let $ X, Y $ be independent with characteristic functions $varphi_X(t) , varphi_Y(t) $.
Show that: $$ sup_{t in mathbb{R}}mid varphi_x(t) - varphi_Y(t) mid le 2P(X neq Y) $$
I would appreciate any tips or hints.
probability random-variables
asked Dec 17 at 16:27
Wywana
535
add a comment |
Let $ X, Y $ be independent with characteristic functions $varphi_X(t) , varphi_Y(t) $.
Show that: $$ sup_{t in mathbb{R}}mid varphi_x(t) - varphi_Y(t) mid le 2P(X neq Y) $$
I would appreciate any tips or hints.
probability random-variables
asked Dec 17 at 16:27
Wywana
535
That's a nice question. Looking forward to an answer.
– Hendrra
Dec 17 at 16:47
add a comment |
Let $ X, Y $ be independent with characteristic functions $varphi_X(t) , varphi_Y(t) $.
Show that: $$ sup_{t in mathbb{R}}mid varphi_x(t) - varphi_Y(t) mid le 2P(X neq Y) $$
I would appreciate any tips or hints.
probability random-variables
asked Dec 17 at 16:27
Wywana
535
Let $ X, Y $ be independent with characteristic functions $varphi_X(t) , varphi_Y(t) $.
Show that: $$ sup_{t in mathbb{R}}mid varphi_x(t) - varphi_Y(t) mid le 2P(X neq Y) $$
I would appreciate any tips or hints.
probability random-variables
probability random-variables
asked Dec 17 at 16:27
Wywana
535
asked Dec 17 at 16:27
Wywana
535
edited Dec 17 at 16:33
asked Dec 17 at 16:27
Wywana
535
asked Dec 17 at 16:27
Wywana
535
asked Dec 17 at 16:27
Wywana
535
535
That's a nice question. Looking forward to an answer.
– Hendrra
Dec 17 at 16:47
add a comment |
That's a nice question. Looking forward to an answer.
– Hendrra
Dec 17 at 16:47
That's a nice question. Looking forward to an answer.
– Hendrra
Dec 17 at 16:47
That's a nice question. Looking forward to an answer.
– Hendrra
Dec 17 at 16:47
add a comment |
2 Answers
2
active
oldest
votes
Note that for all $t,x,yin mathbb{R}$, it holds
$$
|e^{itx}-e^{ity}|leq 2 cdot 1_{{xneq y}}.
$$ Thus, for any $tin mathbb{R}$, we have
$$
|varphi_X(t)-varphi_Y(t)|leq E[|e^{itX}-e^{itY}|]leq 2E[1_{{Xneq Y}}]=2P(Xneq Y).
$$ Now, take supremum over $tin mathbb{R}$ to get
$$
sup_{tinmathbb{R}}|varphi_X(t)-varphi_Y(t)|leq 2P(Xneq Y),
$$as desired.
answered Dec 17 at 18:29
Song
3,985316
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
add a comment |
What sort of distributions do $X$ and $Y$ have? If the distributions have no discrete terms, $P(Xne Y)=1$ and the inequality always hold trivially, since all $|phi(t)|le 1.$
answered Dec 17 at 18:15
herb steinberg
2,4682310
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that for all $t,x,yin mathbb{R}$, it holds
$$
|e^{itx}-e^{ity}|leq 2 cdot 1_{{xneq y}}.
$$ Thus, for any $tin mathbb{R}$, we have
$$
|varphi_X(t)-varphi_Y(t)|leq E[|e^{itX}-e^{itY}|]leq 2E[1_{{Xneq Y}}]=2P(Xneq Y).
$$ Now, take supremum over $tin mathbb{R}$ to get
$$
sup_{tinmathbb{R}}|varphi_X(t)-varphi_Y(t)|leq 2P(Xneq Y),
$$as desired.
answered Dec 17 at 18:29
Song
3,985316
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
add a comment |
Note that for all $t,x,yin mathbb{R}$, it holds
$$
|e^{itx}-e^{ity}|leq 2 cdot 1_{{xneq y}}.
$$ Thus, for any $tin mathbb{R}$, we have
$$
|varphi_X(t)-varphi_Y(t)|leq E[|e^{itX}-e^{itY}|]leq 2E[1_{{Xneq Y}}]=2P(Xneq Y).
$$ Now, take supremum over $tin mathbb{R}$ to get
$$
sup_{tinmathbb{R}}|varphi_X(t)-varphi_Y(t)|leq 2P(Xneq Y),
$$as desired.
answered Dec 17 at 18:29
Song
3,985316
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
add a comment |
Note that for all $t,x,yin mathbb{R}$, it holds
$$
|e^{itx}-e^{ity}|leq 2 cdot 1_{{xneq y}}.
$$ Thus, for any $tin mathbb{R}$, we have
$$
|varphi_X(t)-varphi_Y(t)|leq E[|e^{itX}-e^{itY}|]leq 2E[1_{{Xneq Y}}]=2P(Xneq Y).
$$ Now, take supremum over $tin mathbb{R}$ to get
$$
sup_{tinmathbb{R}}|varphi_X(t)-varphi_Y(t)|leq 2P(Xneq Y),
$$as desired.
answered Dec 17 at 18:29
Song
3,985316
Note that for all $t,x,yin mathbb{R}$, it holds
$$
|e^{itx}-e^{ity}|leq 2 cdot 1_{{xneq y}}.
$$ Thus, for any $tin mathbb{R}$, we have
$$
|varphi_X(t)-varphi_Y(t)|leq E[|e^{itX}-e^{itY}|]leq 2E[1_{{Xneq Y}}]=2P(Xneq Y).
$$ Now, take supremum over $tin mathbb{R}$ to get
$$
sup_{tinmathbb{R}}|varphi_X(t)-varphi_Y(t)|leq 2P(Xneq Y),
$$as desired.
answered Dec 17 at 18:29
Song
3,985316
answered Dec 17 at 18:29
Song
3,985316
answered Dec 17 at 18:29
Song
3,985316
answered Dec 17 at 18:29
Song
3,985316
3,985316
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
add a comment |
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
That's a tidy solution!
– Hendrra
Dec 17 at 19:34
add a comment |
What sort of distributions do $X$ and $Y$ have? If the distributions have no discrete terms, $P(Xne Y)=1$ and the inequality always hold trivially, since all $|phi(t)|le 1.$
answered Dec 17 at 18:15
herb steinberg
2,4682310
add a comment |
What sort of distributions do $X$ and $Y$ have? If the distributions have no discrete terms, $P(Xne Y)=1$ and the inequality always hold trivially, since all $|phi(t)|le 1.$
answered Dec 17 at 18:15
herb steinberg
2,4682310
add a comment |
What sort of distributions do $X$ and $Y$ have? If the distributions have no discrete terms, $P(Xne Y)=1$ and the inequality always hold trivially, since all $|phi(t)|le 1.$
answered Dec 17 at 18:15
herb steinberg
2,4682310
What sort of distributions do $X$ and $Y$ have? If the distributions have no discrete terms, $P(Xne Y)=1$ and the inequality always hold trivially, since all $|phi(t)|le 1.$
answered Dec 17 at 18:15
herb steinberg
2,4682310
answered Dec 17 at 18:15
herb steinberg
2,4682310
answered Dec 17 at 18:15
herb steinberg
2,4682310
answered Dec 17 at 18:15
herb steinberg
2,4682310
2,4682310
add a comment |
add a comment |
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That's a nice question. Looking forward to an answer.
– Hendrra
Dec 17 at 16:47