Mixing
Let $Xsim Poi(10) $ and $Y sim exp(frac{1}{10})$ independent. Why $P(X+Yleqfrac{3}{2}) = P(X=0, Y leq...
$begingroup$
Let $Xsim Poi(10) $ and $Y sim exp(frac{1}{10})$ independent random raviables
I would like to compute:
$P(X+Yleqfrac{3}{2})$
So what I did, which is probably wrong, is the following:
Let $X=k$ then $P(X+Yleqfrac{3}{2})=P(X=k, Yleqfrac{3}{2}-k)$
But I don't know how to take it from there.
The solution on the other hand computes it as follows:
$P(X+Yleqfrac{3}{2}) = P(X=0, Y leq frac{3}{2}) + P(X=1,Yleq frac {1}{2})$
But why does $Xin${0,1} ? it doesn't make much sense to me.
Thanks!!
probability poisson-distribution exponential-distribution
asked Jan 30 at 9:50
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Let $Xsim Poi(10) $ and $Y sim exp(frac{1}{10})$ independent random raviables
I would like to compute:
$P(X+Yleqfrac{3}{2})$
So what I did, which is probably wrong, is the following:
Let $X=k$ then $P(X+Yleqfrac{3}{2})=P(X=k, Yleqfrac{3}{2}-k)$
But I don't know how to take it from there.
The solution on the other hand computes it as follows:
$P(X+Yleqfrac{3}{2}) = P(X=0, Y leq frac{3}{2}) + P(X=1,Yleq frac {1}{2})$
But why does $Xin${0,1} ? it doesn't make much sense to me.
Thanks!!
probability poisson-distribution exponential-distribution
asked Jan 30 at 9:50
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
Let $Xsim Poi(10) $ and $Y sim exp(frac{1}{10})$ independent random raviables
I would like to compute:
$P(X+Yleqfrac{3}{2})$
So what I did, which is probably wrong, is the following:
Let $X=k$ then $P(X+Yleqfrac{3}{2})=P(X=k, Yleqfrac{3}{2}-k)$
But I don't know how to take it from there.
The solution on the other hand computes it as follows:
$P(X+Yleqfrac{3}{2}) = P(X=0, Y leq frac{3}{2}) + P(X=1,Yleq frac {1}{2})$
But why does $Xin${0,1} ? it doesn't make much sense to me.
Thanks!!
probability poisson-distribution exponential-distribution
asked Jan 30 at 9:50
superuser123superuser123
48628
$endgroup$
Let $Xsim Poi(10) $ and $Y sim exp(frac{1}{10})$ independent random raviables
I would like to compute:
$P(X+Yleqfrac{3}{2})$
So what I did, which is probably wrong, is the following:
Let $X=k$ then $P(X+Yleqfrac{3}{2})=P(X=k, Yleqfrac{3}{2}-k)$
But I don't know how to take it from there.
The solution on the other hand computes it as follows:
$P(X+Yleqfrac{3}{2}) = P(X=0, Y leq frac{3}{2}) + P(X=1,Yleq frac {1}{2})$
But why does $Xin${0,1} ? it doesn't make much sense to me.
Thanks!!
probability poisson-distribution exponential-distribution
probability poisson-distribution exponential-distribution
asked Jan 30 at 9:50
superuser123superuser123
48628
asked Jan 30 at 9:50
superuser123superuser123
48628
asked Jan 30 at 9:50
superuser123superuser123
48628
asked Jan 30 at 9:50
superuser123superuser123
48628
asked Jan 30 at 9:50
superuser123superuser123
48628
48628
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$X$ and $Y$ take only non-negative integer values. $X+Y leq frac 3 2$ implies $Xleq frac 3 2$ and the only integers less than or equal to $frac 3 2$ are $0$ and $1$.
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
add a comment |
$begingroup$
That's because $X in mathbb{N} cup {0}$ and $Y > 0$, so $X + Y$ can only be below $3/2$ when $X = 0$ or $X = 1$.
answered Jan 30 at 9:57
HarnakHarnak
1,334512
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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$begingroup$
$X$ and $Y$ take only non-negative integer values. $X+Y leq frac 3 2$ implies $Xleq frac 3 2$ and the only integers less than or equal to $frac 3 2$ are $0$ and $1$.
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
add a comment |
$begingroup$
$X$ and $Y$ take only non-negative integer values. $X+Y leq frac 3 2$ implies $Xleq frac 3 2$ and the only integers less than or equal to $frac 3 2$ are $0$ and $1$.
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
add a comment |
$begingroup$
$X$ and $Y$ take only non-negative integer values. $X+Y leq frac 3 2$ implies $Xleq frac 3 2$ and the only integers less than or equal to $frac 3 2$ are $0$ and $1$.
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
$endgroup$
$X$ and $Y$ take only non-negative integer values. $X+Y leq frac 3 2$ implies $Xleq frac 3 2$ and the only integers less than or equal to $frac 3 2$ are $0$ and $1$.
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
answered Jan 30 at 9:56
Kavi Rama MurthyKavi Rama Murthy
71.9k53170
71.9k53170
add a comment |
add a comment |
$begingroup$
That's because $X in mathbb{N} cup {0}$ and $Y > 0$, so $X + Y$ can only be below $3/2$ when $X = 0$ or $X = 1$.
answered Jan 30 at 9:57
HarnakHarnak
1,334512
$endgroup$
add a comment |
$begingroup$
That's because $X in mathbb{N} cup {0}$ and $Y > 0$, so $X + Y$ can only be below $3/2$ when $X = 0$ or $X = 1$.
answered Jan 30 at 9:57
HarnakHarnak
1,334512
$endgroup$
add a comment |
$begingroup$
That's because $X in mathbb{N} cup {0}$ and $Y > 0$, so $X + Y$ can only be below $3/2$ when $X = 0$ or $X = 1$.
answered Jan 30 at 9:57
HarnakHarnak
1,334512
$endgroup$
That's because $X in mathbb{N} cup {0}$ and $Y > 0$, so $X + Y$ can only be below $3/2$ when $X = 0$ or $X = 1$.
answered Jan 30 at 9:57
HarnakHarnak
1,334512
answered Jan 30 at 9:57
HarnakHarnak
1,334512
answered Jan 30 at 9:57
HarnakHarnak
1,334512
answered Jan 30 at 9:57
HarnakHarnak
1,334512
1,334512
add a comment |
add a comment |
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