Mixing
Sum of absolutely continuous independent random variables
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I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).
Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
$mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$
How did we get this last step?
random-variables
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
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add a comment |
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I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).
Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
$mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$
How did we get this last step?
random-variables
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
$endgroup$
add a comment |
$begingroup$
I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).
Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
$mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$
How did we get this last step?
random-variables
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
$endgroup$
I'm facing the proof of a theorem stating that the sum of two absolutely continuous independent random variables is a new absolutely continuous random variable whose density is given by the convolution of the density functions of the variables. At the beginning I find this statement that I'm struggling to understand (consider that before this theorem we onlybhad the definition of A.C. random variable and independence of random variables).
Let $X_1, X_2$ be two absolutely continuous random variables, and $rho_1(t), rho_2(t)$ be their density. Let $Y=X_1+X_2$. Then
$mathbb{P}( Y leq t )= mathbb{P}( X_1+ X_2 leq t ) = intlimits_{-infty}^{+infty} rho_1(t_1) cdot intlimits_{-infty}^{t-t_1} rho_2(t_2) dt_2 dt_1$
How did we get this last step?
random-variables
random-variables
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
asked Jan 2 at 9:47
Baffo rastaBaffo rasta
10511
10511
add a comment |
add a comment |
1 Answer
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$$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
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1 Answer
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1 Answer
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active
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$begingroup$
$$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
add a comment |
$begingroup$
$$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
add a comment |
$begingroup$
$$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
$endgroup$
$$P{X_1+X_2 leq t|X_1=t_1)=P{X_2 leq t -X_1|X_1=t_1)$$ $=int_{-infty} ^{t-t_1} rho_2(t_2)dt_2.$ Now integrate w.r.t the distribution of $X_1$.
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
answered Jan 2 at 12:14
Kavi Rama MurthyKavi Rama Murthy
53k32055
53k32055
add a comment |
add a comment |
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