Calculating a density=1
$begingroup$
I had an excercise regarding a family of densities. My question is: Is this really a family of densities? I calculated the integral over $R^2$ and think that it is only a density when a=1, but I might be wrong. Let $$g(x,y)=frac{1}{4}e^{-a|x|-a^{-1}|y|}$$ so I calculated the integral over $R^{2}$ and determined that is is one iff a=1, can anyone defy? $ain(0,infty)$
probability probability-distributions
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
$endgroup$
add a comment |
$begingroup$
I had an excercise regarding a family of densities. My question is: Is this really a family of densities? I calculated the integral over $R^2$ and think that it is only a density when a=1, but I might be wrong. Let $$g(x,y)=frac{1}{4}e^{-a|x|-a^{-1}|y|}$$ so I calculated the integral over $R^{2}$ and determined that is is one iff a=1, can anyone defy? $ain(0,infty)$
probability probability-distributions
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
$endgroup$
$begingroup$
If density in this context means probability density, then the integral must be $1$.
$endgroup$
– herb steinberg
Dec 7 '18 at 18:42
$begingroup$
I know, but according to my calculation it is one only if a=1, but should be for all a in $(0,infty)$
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:44
$begingroup$
Ech, I think you may be right, but they should have written it... i lost time and points because of that on my exam. Precisely, it is written let the probability measure $mu_a$ be on $R^{2}$ with the density I have written.
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:51
$begingroup$
"so I calculated the integral over R2 and determined that is is one iff a=1" We cannot know where you erred since you do not show what you did...
$endgroup$
– Did
Dec 7 '18 at 19:07
$begingroup$
I already know my mistake thanks, I was so tired that I added $frac{a}{2}+frac{2}{a}$ instead of multiplying them
$endgroup$
– ryszard eggink
Dec 7 '18 at 19:30
add a comment |
$begingroup$
I had an excercise regarding a family of densities. My question is: Is this really a family of densities? I calculated the integral over $R^2$ and think that it is only a density when a=1, but I might be wrong. Let $$g(x,y)=frac{1}{4}e^{-a|x|-a^{-1}|y|}$$ so I calculated the integral over $R^{2}$ and determined that is is one iff a=1, can anyone defy? $ain(0,infty)$
probability probability-distributions
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
$endgroup$
I had an excercise regarding a family of densities. My question is: Is this really a family of densities? I calculated the integral over $R^2$ and think that it is only a density when a=1, but I might be wrong. Let $$g(x,y)=frac{1}{4}e^{-a|x|-a^{-1}|y|}$$ so I calculated the integral over $R^{2}$ and determined that is is one iff a=1, can anyone defy? $ain(0,infty)$
probability probability-distributions
probability probability-distributions
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
asked Dec 7 '18 at 18:39
ryszard egginkryszard eggink
314110
314110
$begingroup$
If density in this context means probability density, then the integral must be $1$.
$endgroup$
– herb steinberg
Dec 7 '18 at 18:42
$begingroup$
I know, but according to my calculation it is one only if a=1, but should be for all a in $(0,infty)$
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:44
$begingroup$
Ech, I think you may be right, but they should have written it... i lost time and points because of that on my exam. Precisely, it is written let the probability measure $mu_a$ be on $R^{2}$ with the density I have written.
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:51
$begingroup$
"so I calculated the integral over R2 and determined that is is one iff a=1" We cannot know where you erred since you do not show what you did...
$endgroup$
– Did
Dec 7 '18 at 19:07
$begingroup$
I already know my mistake thanks, I was so tired that I added $frac{a}{2}+frac{2}{a}$ instead of multiplying them
$endgroup$
– ryszard eggink
Dec 7 '18 at 19:30
add a comment |
$begingroup$
If density in this context means probability density, then the integral must be $1$.
$endgroup$
– herb steinberg
Dec 7 '18 at 18:42
$begingroup$
I know, but according to my calculation it is one only if a=1, but should be for all a in $(0,infty)$
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:44
$begingroup$
Ech, I think you may be right, but they should have written it... i lost time and points because of that on my exam. Precisely, it is written let the probability measure $mu_a$ be on $R^{2}$ with the density I have written.
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:51
$begingroup$
"so I calculated the integral over R2 and determined that is is one iff a=1" We cannot know where you erred since you do not show what you did...
$endgroup$
– Did
Dec 7 '18 at 19:07
$begingroup$
I already know my mistake thanks, I was so tired that I added $frac{a}{2}+frac{2}{a}$ instead of multiplying them
$endgroup$
– ryszard eggink
Dec 7 '18 at 19:30
$begingroup$
If density in this context means probability density, then the integral must be $1$.
$endgroup$
– herb steinberg
Dec 7 '18 at 18:42
$begingroup$
If density in this context means probability density, then the integral must be $1$.
$endgroup$
– herb steinberg
Dec 7 '18 at 18:42
$begingroup$
I know, but according to my calculation it is one only if a=1, but should be for all a in $(0,infty)$
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:44
$begingroup$
I know, but according to my calculation it is one only if a=1, but should be for all a in $(0,infty)$
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:44
$begingroup$
Ech, I think you may be right, but they should have written it... i lost time and points because of that on my exam. Precisely, it is written let the probability measure $mu_a$ be on $R^{2}$ with the density I have written.
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:51
$begingroup$
Ech, I think you may be right, but they should have written it... i lost time and points because of that on my exam. Precisely, it is written let the probability measure $mu_a$ be on $R^{2}$ with the density I have written.
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:51
$begingroup$
"so I calculated the integral over R2 and determined that is is one iff a=1" We cannot know where you erred since you do not show what you did...
$endgroup$
– Did
Dec 7 '18 at 19:07
$begingroup$
"so I calculated the integral over R2 and determined that is is one iff a=1" We cannot know where you erred since you do not show what you did...
$endgroup$
– Did
Dec 7 '18 at 19:07
$begingroup$
I already know my mistake thanks, I was so tired that I added $frac{a}{2}+frac{2}{a}$ instead of multiplying them
$endgroup$
– ryszard eggink
Dec 7 '18 at 19:30
$begingroup$
I already know my mistake thanks, I was so tired that I added $frac{a}{2}+frac{2}{a}$ instead of multiplying them
$endgroup$
– ryszard eggink
Dec 7 '18 at 19:30
add a comment |
1 Answer
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$int_{R^2} g(x,y)dxdy=(frac{1}{2}int_R e^{-a^{-1}|y|}dy)(frac{1}{2}int_R e^{-a|x|}dx)={(a)(frac{1}{a})=1}$,
for all $agt 0$.
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
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add a comment |
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$begingroup$
$int_{R^2} g(x,y)dxdy=(frac{1}{2}int_R e^{-a^{-1}|y|}dy)(frac{1}{2}int_R e^{-a|x|}dx)={(a)(frac{1}{a})=1}$,
for all $agt 0$.
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
$endgroup$
add a comment |
$begingroup$
$int_{R^2} g(x,y)dxdy=(frac{1}{2}int_R e^{-a^{-1}|y|}dy)(frac{1}{2}int_R e^{-a|x|}dx)={(a)(frac{1}{a})=1}$,
for all $agt 0$.
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
$endgroup$
add a comment |
$begingroup$
$int_{R^2} g(x,y)dxdy=(frac{1}{2}int_R e^{-a^{-1}|y|}dy)(frac{1}{2}int_R e^{-a|x|}dx)={(a)(frac{1}{a})=1}$,
for all $agt 0$.
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
$endgroup$
$int_{R^2} g(x,y)dxdy=(frac{1}{2}int_R e^{-a^{-1}|y|}dy)(frac{1}{2}int_R e^{-a|x|}dx)={(a)(frac{1}{a})=1}$,
for all $agt 0$.
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
edited Dec 7 '18 at 20:09
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
answered Dec 7 '18 at 18:57
herb steinbergherb steinberg
2,5932310
2,5932310
add a comment |
add a comment |
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$begingroup$
If density in this context means probability density, then the integral must be $1$.
$endgroup$
– herb steinberg
Dec 7 '18 at 18:42
$begingroup$
I know, but according to my calculation it is one only if a=1, but should be for all a in $(0,infty)$
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:44
$begingroup$
Ech, I think you may be right, but they should have written it... i lost time and points because of that on my exam. Precisely, it is written let the probability measure $mu_a$ be on $R^{2}$ with the density I have written.
$endgroup$
– ryszard eggink
Dec 7 '18 at 18:51
$begingroup$
"so I calculated the integral over R2 and determined that is is one iff a=1" We cannot know where you erred since you do not show what you did...
$endgroup$
– Did
Dec 7 '18 at 19:07
$begingroup$
I already know my mistake thanks, I was so tired that I added $frac{a}{2}+frac{2}{a}$ instead of multiplying them
$endgroup$
– ryszard eggink
Dec 7 '18 at 19:30