joint cumulative distribution function proof
$begingroup$
Let $X=(X_1,..,X_n)$ be a density random vector.
Then for all $1 le k le n$: $$f_{X_k}(x_k)=int_{-infty}^{infty}dx_1···int_{-infty}^{infty}dx_{k-1}int_{-infty}^{infty}dx_{k+1}···int_{-infty}^{infty}dx_n f_X(x_1,..,x_n)$$
Proof:
$$P(X_kin A)=P(Xinmathbb{R}^{k-1}times A times mathbb{R}^{n-k})=$$$$int_{mathbb{R}}dx_1···int_{mathbb{R}}dx_{k-1}int_{A}dx_kint_{mathbb{R}}dx_{k+1}int_{mathbb{R}}dx_{n}f_X(x_1,..,x_n)=$$$$int_{A}dx_kf_{X_k}(x_k)$$
I don't understand the last two equalities, is it by definition that $P(X_1le x_1, ...,X_n le x_n)=int_{-infty}^{x_1}ds_1···int_{-infty}^{x_n}ds_nf_X(s_1,..,s_n)$ ?
And how to explain the last equality ?
probability-theory density-function
edited Dec 19 '18 at 10:14
drhab
102k545136
asked Dec 19 '18 at 9:02
WaspWasp
112
$endgroup$
add a comment |
$begingroup$
Let $X=(X_1,..,X_n)$ be a density random vector.
Then for all $1 le k le n$: $$f_{X_k}(x_k)=int_{-infty}^{infty}dx_1···int_{-infty}^{infty}dx_{k-1}int_{-infty}^{infty}dx_{k+1}···int_{-infty}^{infty}dx_n f_X(x_1,..,x_n)$$
Proof:
$$P(X_kin A)=P(Xinmathbb{R}^{k-1}times A times mathbb{R}^{n-k})=$$$$int_{mathbb{R}}dx_1···int_{mathbb{R}}dx_{k-1}int_{A}dx_kint_{mathbb{R}}dx_{k+1}int_{mathbb{R}}dx_{n}f_X(x_1,..,x_n)=$$$$int_{A}dx_kf_{X_k}(x_k)$$
I don't understand the last two equalities, is it by definition that $P(X_1le x_1, ...,X_n le x_n)=int_{-infty}^{x_1}ds_1···int_{-infty}^{x_n}ds_nf_X(s_1,..,s_n)$ ?
And how to explain the last equality ?
probability-theory density-function
edited Dec 19 '18 at 10:14
drhab
102k545136
asked Dec 19 '18 at 9:02
WaspWasp
112
$endgroup$
add a comment |
$begingroup$
Let $X=(X_1,..,X_n)$ be a density random vector.
Then for all $1 le k le n$: $$f_{X_k}(x_k)=int_{-infty}^{infty}dx_1···int_{-infty}^{infty}dx_{k-1}int_{-infty}^{infty}dx_{k+1}···int_{-infty}^{infty}dx_n f_X(x_1,..,x_n)$$
Proof:
$$P(X_kin A)=P(Xinmathbb{R}^{k-1}times A times mathbb{R}^{n-k})=$$$$int_{mathbb{R}}dx_1···int_{mathbb{R}}dx_{k-1}int_{A}dx_kint_{mathbb{R}}dx_{k+1}int_{mathbb{R}}dx_{n}f_X(x_1,..,x_n)=$$$$int_{A}dx_kf_{X_k}(x_k)$$
I don't understand the last two equalities, is it by definition that $P(X_1le x_1, ...,X_n le x_n)=int_{-infty}^{x_1}ds_1···int_{-infty}^{x_n}ds_nf_X(s_1,..,s_n)$ ?
And how to explain the last equality ?
probability-theory density-function
edited Dec 19 '18 at 10:14
drhab
102k545136
asked Dec 19 '18 at 9:02
WaspWasp
112
$endgroup$
Let $X=(X_1,..,X_n)$ be a density random vector.
Then for all $1 le k le n$: $$f_{X_k}(x_k)=int_{-infty}^{infty}dx_1···int_{-infty}^{infty}dx_{k-1}int_{-infty}^{infty}dx_{k+1}···int_{-infty}^{infty}dx_n f_X(x_1,..,x_n)$$
Proof:
$$P(X_kin A)=P(Xinmathbb{R}^{k-1}times A times mathbb{R}^{n-k})=$$$$int_{mathbb{R}}dx_1···int_{mathbb{R}}dx_{k-1}int_{A}dx_kint_{mathbb{R}}dx_{k+1}int_{mathbb{R}}dx_{n}f_X(x_1,..,x_n)=$$$$int_{A}dx_kf_{X_k}(x_k)$$
I don't understand the last two equalities, is it by definition that $P(X_1le x_1, ...,X_n le x_n)=int_{-infty}^{x_1}ds_1···int_{-infty}^{x_n}ds_nf_X(s_1,..,s_n)$ ?
And how to explain the last equality ?
probability-theory density-function
probability-theory density-function
edited Dec 19 '18 at 10:14
drhab
102k545136
asked Dec 19 '18 at 9:02
WaspWasp
112
edited Dec 19 '18 at 10:14
drhab
102k545136
asked Dec 19 '18 at 9:02
WaspWasp
112
edited Dec 19 '18 at 10:14
drhab
102k545136
edited Dec 19 '18 at 10:14
drhab
102k545136
edited Dec 19 '18 at 10:14
drhab
102k545136
102k545136
asked Dec 19 '18 at 9:02
WaspWasp
112
asked Dec 19 '18 at 9:02
WaspWasp
112
asked Dec 19 '18 at 9:02
WaspWasp
112
112
add a comment |
add a comment |
1 Answer
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$begingroup$
It is by definition that $$P((X_1,dots,X_n)in B)=int;d(x_1,dots,x_n)mathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
or in slightly different notation:$$P((X_1,dots,X_n)in B)=int_B;d(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
As a sidenote: personally I dislike this notation and would rather go for $$P((X_1,dots,X_n)in B)=intdotsintmathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n);dx_1dots dx_n$$
In special case $B=A_1timescdotscdotstimes A_n$ this takes the form:$$P(X_1in A_1,dots,X_nin A_n)=int_{A_1};dx_1cdotsint_{A_n};dx_nf_X(x_1,dots,x_n)$$
Now see what happens if $A_i=(-infty,x_i]$.
Concerning the second equality note that it is allowed here to change the order of integration.
So: $$int_{mathbb R}dx_1cdotsint_Adx_kcdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_{A}dx_kint_{mathbb R}dx_1cdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_Adx_kf_{X_k}(x_k)$$
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
$endgroup$
add a comment |
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1 Answer
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$begingroup$
It is by definition that $$P((X_1,dots,X_n)in B)=int;d(x_1,dots,x_n)mathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
or in slightly different notation:$$P((X_1,dots,X_n)in B)=int_B;d(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
As a sidenote: personally I dislike this notation and would rather go for $$P((X_1,dots,X_n)in B)=intdotsintmathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n);dx_1dots dx_n$$
In special case $B=A_1timescdotscdotstimes A_n$ this takes the form:$$P(X_1in A_1,dots,X_nin A_n)=int_{A_1};dx_1cdotsint_{A_n};dx_nf_X(x_1,dots,x_n)$$
Now see what happens if $A_i=(-infty,x_i]$.
Concerning the second equality note that it is allowed here to change the order of integration.
So: $$int_{mathbb R}dx_1cdotsint_Adx_kcdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_{A}dx_kint_{mathbb R}dx_1cdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_Adx_kf_{X_k}(x_k)$$
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
It is by definition that $$P((X_1,dots,X_n)in B)=int;d(x_1,dots,x_n)mathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
or in slightly different notation:$$P((X_1,dots,X_n)in B)=int_B;d(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
As a sidenote: personally I dislike this notation and would rather go for $$P((X_1,dots,X_n)in B)=intdotsintmathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n);dx_1dots dx_n$$
In special case $B=A_1timescdotscdotstimes A_n$ this takes the form:$$P(X_1in A_1,dots,X_nin A_n)=int_{A_1};dx_1cdotsint_{A_n};dx_nf_X(x_1,dots,x_n)$$
Now see what happens if $A_i=(-infty,x_i]$.
Concerning the second equality note that it is allowed here to change the order of integration.
So: $$int_{mathbb R}dx_1cdotsint_Adx_kcdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_{A}dx_kint_{mathbb R}dx_1cdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_Adx_kf_{X_k}(x_k)$$
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
It is by definition that $$P((X_1,dots,X_n)in B)=int;d(x_1,dots,x_n)mathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
or in slightly different notation:$$P((X_1,dots,X_n)in B)=int_B;d(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
As a sidenote: personally I dislike this notation and would rather go for $$P((X_1,dots,X_n)in B)=intdotsintmathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n);dx_1dots dx_n$$
In special case $B=A_1timescdotscdotstimes A_n$ this takes the form:$$P(X_1in A_1,dots,X_nin A_n)=int_{A_1};dx_1cdotsint_{A_n};dx_nf_X(x_1,dots,x_n)$$
Now see what happens if $A_i=(-infty,x_i]$.
Concerning the second equality note that it is allowed here to change the order of integration.
So: $$int_{mathbb R}dx_1cdotsint_Adx_kcdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_{A}dx_kint_{mathbb R}dx_1cdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_Adx_kf_{X_k}(x_k)$$
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
$endgroup$
It is by definition that $$P((X_1,dots,X_n)in B)=int;d(x_1,dots,x_n)mathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
or in slightly different notation:$$P((X_1,dots,X_n)in B)=int_B;d(x_1,dots,x_n)f_X(x_1,dots,x_n)$$
As a sidenote: personally I dislike this notation and would rather go for $$P((X_1,dots,X_n)in B)=intdotsintmathbf1_B(x_1,dots,x_n)f_X(x_1,dots,x_n);dx_1dots dx_n$$
In special case $B=A_1timescdotscdotstimes A_n$ this takes the form:$$P(X_1in A_1,dots,X_nin A_n)=int_{A_1};dx_1cdotsint_{A_n};dx_nf_X(x_1,dots,x_n)$$
Now see what happens if $A_i=(-infty,x_i]$.
Concerning the second equality note that it is allowed here to change the order of integration.
So: $$int_{mathbb R}dx_1cdotsint_Adx_kcdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_{A}dx_kint_{mathbb R}dx_1cdotsint_{mathbb R}dx_nf_X(x_1,dots,x_n)=int_Adx_kf_{X_k}(x_k)$$
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
edited Dec 19 '18 at 17:32
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
answered Dec 19 '18 at 10:43
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
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