$X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $1$ and...
$begingroup$
Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$
My input
$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $
$X_1sim Exp(1)=e^{-x}implies G(1,1)$
Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $
Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$
$X_2sim G(1,2)$
$P(X_2-X_1<0)$
I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.
probability statistics probability-distributions
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
$endgroup$
add a comment |
$begingroup$
Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$
My input
$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $
$X_1sim Exp(1)=e^{-x}implies G(1,1)$
Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $
Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$
$X_2sim G(1,2)$
$P(X_2-X_1<0)$
I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.
probability statistics probability-distributions
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
$endgroup$
1
$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26
$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27
3
$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29
add a comment |
$begingroup$
Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$
My input
$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $
$X_1sim Exp(1)=e^{-x}implies G(1,1)$
Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $
Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$
$X_2sim G(1,2)$
$P(X_2-X_1<0)$
I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.
probability statistics probability-distributions
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
$endgroup$
Let $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $2$ and variance $2$. If $X_1$ and $X_2$ are Independent random variable then $P(X_1<X_2)$
My input
$Exp(a)=ae^{-ax} ; G(a,lambda)=dfrac{a^{lambda}}{Gamma{lambda} }e^{ax}x^{lambda-1} $
$X_1sim Exp(1)=e^{-x}implies G(1,1)$
Mean $ =dfrac{lambda}{a}=2 implies lambda=2a $
Variance = $dfrac{lambda}{a^2}=dfrac{2a}{a^2}=2 implies a=1,lambda=2$
$X_2sim G(1,2)$
$P(X_2-X_1<0)$
I am trying to find out distribution of $X_2-X_1$ I tried to subtract the parameter($lambda_1,lambda_2$) but the property of MGF works in addition only.
Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $X_2$ so I am out of option I need a hint or something.
probability statistics probability-distributions
probability statistics probability-distributions
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
asked Dec 13 '18 at 14:21
Daman deepDaman deep
756418
756418
1
$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26
$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27
3
$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29
add a comment |
1
$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26
$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27
3
$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29
1
1
$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26
$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26
$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27
$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27
3
3
$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29
$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29
add a comment |
1 Answer
1
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$begingroup$
Smiley is right.
Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$
(and leave $X_2-X_1$ for what it is!)
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Smiley is right.
Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$
(and leave $X_2-X_1$ for what it is!)
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
Smiley is right.
Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$
(and leave $X_2-X_1$ for what it is!)
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
Smiley is right.
Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$
(and leave $X_2-X_1$ for what it is!)
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
$endgroup$
Smiley is right.
Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1geq X_2)=1-int_0^{infty} P(X_1geq x)f_{X_2}(x)dx=1-int_0^{infty} e^{-x}f_{X_2}(x)dx$$
(and leave $X_2-X_1$ for what it is!)
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
answered Dec 13 '18 at 14:42
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
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1
$begingroup$
If you fix $X_2$, can you find $P(X_1<X_2)$?
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:26
$begingroup$
But won't it include $X_2$ how will I get final result ?
$endgroup$
– Daman deep
Dec 13 '18 at 14:27
3
$begingroup$
If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $int P(X_1<X_2)f_G(X_2)mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer.
$endgroup$
– SmileyCraft
Dec 13 '18 at 14:29