Mixing
Probability of an event for a continuous random vector of three coordinates
$begingroup$
Let $X=(X_1,X_2,X_3)$ be a continuous random vector of the joint pdf $$f(x_1,x_2,x_3)= 12x_2 ;mathrm f mathrm o mathrm r ; 0<x_3<x_2<x_1<1$$ and $0$ elsewhere.
I need to find the probability of event $B$ where $B={x_3leq1/3}$. I've been able to sketch the support and see that it's a sort of pyramidal shape, and know that I should be able to find the probability via a triple integral.
My instinct says to simply set it up like so:
$$int_0^{1/3}int_{x_3}^{x_1}int_{x_2}^{1}12x_2;dx_1dx_2dx_3$$
but this is clearly wrong since the answer will still be in terms of a variable. If anyone could help me understand where I've been going wrong it would be greatly appreciated!
probability probability-distributions
asked Jan 20 at 20:25
sk13sk13
246
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add a comment |
$begingroup$
Let $X=(X_1,X_2,X_3)$ be a continuous random vector of the joint pdf $$f(x_1,x_2,x_3)= 12x_2 ;mathrm f mathrm o mathrm r ; 0<x_3<x_2<x_1<1$$ and $0$ elsewhere.
I need to find the probability of event $B$ where $B={x_3leq1/3}$. I've been able to sketch the support and see that it's a sort of pyramidal shape, and know that I should be able to find the probability via a triple integral.
My instinct says to simply set it up like so:
$$int_0^{1/3}int_{x_3}^{x_1}int_{x_2}^{1}12x_2;dx_1dx_2dx_3$$
but this is clearly wrong since the answer will still be in terms of a variable. If anyone could help me understand where I've been going wrong it would be greatly appreciated!
probability probability-distributions
asked Jan 20 at 20:25
sk13sk13
246
$endgroup$
add a comment |
$begingroup$
Let $X=(X_1,X_2,X_3)$ be a continuous random vector of the joint pdf $$f(x_1,x_2,x_3)= 12x_2 ;mathrm f mathrm o mathrm r ; 0<x_3<x_2<x_1<1$$ and $0$ elsewhere.
I need to find the probability of event $B$ where $B={x_3leq1/3}$. I've been able to sketch the support and see that it's a sort of pyramidal shape, and know that I should be able to find the probability via a triple integral.
My instinct says to simply set it up like so:
$$int_0^{1/3}int_{x_3}^{x_1}int_{x_2}^{1}12x_2;dx_1dx_2dx_3$$
but this is clearly wrong since the answer will still be in terms of a variable. If anyone could help me understand where I've been going wrong it would be greatly appreciated!
probability probability-distributions
asked Jan 20 at 20:25
sk13sk13
246
$endgroup$
Let $X=(X_1,X_2,X_3)$ be a continuous random vector of the joint pdf $$f(x_1,x_2,x_3)= 12x_2 ;mathrm f mathrm o mathrm r ; 0<x_3<x_2<x_1<1$$ and $0$ elsewhere.
I need to find the probability of event $B$ where $B={x_3leq1/3}$. I've been able to sketch the support and see that it's a sort of pyramidal shape, and know that I should be able to find the probability via a triple integral.
My instinct says to simply set it up like so:
$$int_0^{1/3}int_{x_3}^{x_1}int_{x_2}^{1}12x_2;dx_1dx_2dx_3$$
but this is clearly wrong since the answer will still be in terms of a variable. If anyone could help me understand where I've been going wrong it would be greatly appreciated!
probability probability-distributions
probability probability-distributions
asked Jan 20 at 20:25
sk13sk13
246
asked Jan 20 at 20:25
sk13sk13
246
edited Jan 20 at 21:03
sk13
asked Jan 20 at 20:25
sk13sk13
246
asked Jan 20 at 20:25
sk13sk13
246
asked Jan 20 at 20:25
sk13sk13
246
246
add a comment |
add a comment |
2 Answers
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It is $int_0^{1/3} int_{x_3}^{1} int_{x_2}^{1} 12x_2 dx_1dx_2dx_3$. (The middle integral cannot involve $x_1$. The condition $x_2 <x_1$ is already taken care of in the limits for $x_1$).
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
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add a comment |
$begingroup$
Change the last integral from ($x_2$ to 1) to ($x_1$ to 1). Does it make any sense?
answered Jan 20 at 20:43
ShaqShaq
3049
$endgroup$
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
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
$begingroup$
It is $int_0^{1/3} int_{x_3}^{1} int_{x_2}^{1} 12x_2 dx_1dx_2dx_3$. (The middle integral cannot involve $x_1$. The condition $x_2 <x_1$ is already taken care of in the limits for $x_1$).
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
add a comment |
$begingroup$
It is $int_0^{1/3} int_{x_3}^{1} int_{x_2}^{1} 12x_2 dx_1dx_2dx_3$. (The middle integral cannot involve $x_1$. The condition $x_2 <x_1$ is already taken care of in the limits for $x_1$).
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
add a comment |
$begingroup$
It is $int_0^{1/3} int_{x_3}^{1} int_{x_2}^{1} 12x_2 dx_1dx_2dx_3$. (The middle integral cannot involve $x_1$. The condition $x_2 <x_1$ is already taken care of in the limits for $x_1$).
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
It is $int_0^{1/3} int_{x_3}^{1} int_{x_2}^{1} 12x_2 dx_1dx_2dx_3$. (The middle integral cannot involve $x_1$. The condition $x_2 <x_1$ is already taken care of in the limits for $x_1$).
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
answered Jan 20 at 23:58
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
64.7k42766
add a comment |
add a comment |
$begingroup$
Change the last integral from ($x_2$ to 1) to ($x_1$ to 1). Does it make any sense?
answered Jan 20 at 20:43
ShaqShaq
3049
$endgroup$
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
add a comment |
$begingroup$
Change the last integral from ($x_2$ to 1) to ($x_1$ to 1). Does it make any sense?
answered Jan 20 at 20:43
ShaqShaq
3049
$endgroup$
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
add a comment |
$begingroup$
Change the last integral from ($x_2$ to 1) to ($x_1$ to 1). Does it make any sense?
answered Jan 20 at 20:43
ShaqShaq
3049
$endgroup$
Change the last integral from ($x_2$ to 1) to ($x_1$ to 1). Does it make any sense?
answered Jan 20 at 20:43
ShaqShaq
3049
answered Jan 20 at 20:43
ShaqShaq
3049
answered Jan 20 at 20:43
ShaqShaq
3049
answered Jan 20 at 20:43
ShaqShaq
3049
3049
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
add a comment |
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
$begingroup$
Sorry, I'm not sure I understand. Why would the bounds of $x_1$ be ($x_1$,1)?
$endgroup$
– sk13
Jan 20 at 22:49
add a comment |
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