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Determine the mode of the gamma distribution with parameters $alpha$ and $beta$
2
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How do you determine the mode of a gamma distribution with parameters $alpha$ and $beta$ ? Without looking on Wikipedia.
gamma-distribution
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
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2
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How do you determine the mode of a gamma distribution with parameters $alpha$ and $beta$ ? Without looking on Wikipedia.
gamma-distribution
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
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add a comment |
2
2
2
$begingroup$
How do you determine the mode of a gamma distribution with parameters $alpha$ and $beta$ ? Without looking on Wikipedia.
gamma-distribution
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
$endgroup$
How do you determine the mode of a gamma distribution with parameters $alpha$ and $beta$ ? Without looking on Wikipedia.
gamma-distribution
gamma-distribution
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
asked Mar 30 '17 at 6:15
Amanda R.Amanda R.
14019
14019
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1 Answer
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Hint: you want to maximize $x^{alpha-1} e^{-beta x}$ over $x in (0,infty)$.
The derivative is $e^{-beta x}[(alpha-1)x^{alpha-2} - beta x^{alpha-1}] = x^{alpha-2} e^{-beta x} (alpha-1-beta x)$, which is zero when $x= frac{alpha-1}{beta}$ or $x=0$.
- If $alpha ge 1$, direct inspection shows that $x=0$ is not the mode, since the pdf is zero there. Thus the other critical point $frac{alpha-1}{beta}$ must be the mode.
- If $alpha < 1$ the pdf has a positive asymptote at $x=0$. Moreover the derivative is strictly negative for all $x>0$, so it decreases from $infty$ to $0$ as $x$ goes from $0$ to $infty$.
Thanks to JeanMarie for the clarifications.
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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3
$begingroup$
Hint: you want to maximize $x^{alpha-1} e^{-beta x}$ over $x in (0,infty)$.
The derivative is $e^{-beta x}[(alpha-1)x^{alpha-2} - beta x^{alpha-1}] = x^{alpha-2} e^{-beta x} (alpha-1-beta x)$, which is zero when $x= frac{alpha-1}{beta}$ or $x=0$.
- If $alpha ge 1$, direct inspection shows that $x=0$ is not the mode, since the pdf is zero there. Thus the other critical point $frac{alpha-1}{beta}$ must be the mode.
- If $alpha < 1$ the pdf has a positive asymptote at $x=0$. Moreover the derivative is strictly negative for all $x>0$, so it decreases from $infty$ to $0$ as $x$ goes from $0$ to $infty$.
Thanks to JeanMarie for the clarifications.
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
add a comment |
3
$begingroup$
Hint: you want to maximize $x^{alpha-1} e^{-beta x}$ over $x in (0,infty)$.
The derivative is $e^{-beta x}[(alpha-1)x^{alpha-2} - beta x^{alpha-1}] = x^{alpha-2} e^{-beta x} (alpha-1-beta x)$, which is zero when $x= frac{alpha-1}{beta}$ or $x=0$.
- If $alpha ge 1$, direct inspection shows that $x=0$ is not the mode, since the pdf is zero there. Thus the other critical point $frac{alpha-1}{beta}$ must be the mode.
- If $alpha < 1$ the pdf has a positive asymptote at $x=0$. Moreover the derivative is strictly negative for all $x>0$, so it decreases from $infty$ to $0$ as $x$ goes from $0$ to $infty$.
Thanks to JeanMarie for the clarifications.
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
add a comment |
3
3
3
$begingroup$
Hint: you want to maximize $x^{alpha-1} e^{-beta x}$ over $x in (0,infty)$.
The derivative is $e^{-beta x}[(alpha-1)x^{alpha-2} - beta x^{alpha-1}] = x^{alpha-2} e^{-beta x} (alpha-1-beta x)$, which is zero when $x= frac{alpha-1}{beta}$ or $x=0$.
- If $alpha ge 1$, direct inspection shows that $x=0$ is not the mode, since the pdf is zero there. Thus the other critical point $frac{alpha-1}{beta}$ must be the mode.
- If $alpha < 1$ the pdf has a positive asymptote at $x=0$. Moreover the derivative is strictly negative for all $x>0$, so it decreases from $infty$ to $0$ as $x$ goes from $0$ to $infty$.
Thanks to JeanMarie for the clarifications.
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
$endgroup$
Hint: you want to maximize $x^{alpha-1} e^{-beta x}$ over $x in (0,infty)$.
The derivative is $e^{-beta x}[(alpha-1)x^{alpha-2} - beta x^{alpha-1}] = x^{alpha-2} e^{-beta x} (alpha-1-beta x)$, which is zero when $x= frac{alpha-1}{beta}$ or $x=0$.
- If $alpha ge 1$, direct inspection shows that $x=0$ is not the mode, since the pdf is zero there. Thus the other critical point $frac{alpha-1}{beta}$ must be the mode.
- If $alpha < 1$ the pdf has a positive asymptote at $x=0$. Moreover the derivative is strictly negative for all $x>0$, so it decreases from $infty$ to $0$ as $x$ goes from $0$ to $infty$.
Thanks to JeanMarie for the clarifications.
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
edited Mar 30 '17 at 18:43
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
answered Mar 30 '17 at 6:21
angryavianangryavian
42.5k23481
42.5k23481
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
add a comment |
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
How do we know it is zero when $x=frac{alpha-1}{beta}$ or $x=0$. Zero is not the mode because $x>0$ correct?
$endgroup$
– Amanda R.
Mar 30 '17 at 6:58
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
Look at the curve : It is not 0 because in 0, the pdf is 0 !
$endgroup$
– Jean Marie
Mar 30 '17 at 7:04
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
@JeanMarie can you please explain more
$endgroup$
– Amanda R.
Mar 30 '17 at 7:08
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
You are looking for a mode: it is necessarily a values of $x$ such that the pdf f(x) has a $>0$ value. In $0$ it has a zero value at least for the cases $alpha>1$ (for the other cases, one cannat speak about a mode because the pdf has a vertical asymptote.
$endgroup$
– Jean Marie
Mar 30 '17 at 7:22
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
$begingroup$
@JeanMarie Thanks for the clarifications!
$endgroup$
– angryavian
Mar 30 '17 at 18:43
add a comment |
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