proof if X Weibull distributed then Y=aX^b (a,b>0) also Weibull distributed
$begingroup$
It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:
$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$
= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$
= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$
if k = s-1
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$
So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0
I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.
statistics proof-verification probability-distributions
asked Dec 30 '18 at 9:26
dsamaradsamara
12
$endgroup$
add a comment |
$begingroup$
It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:
$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$
= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$
= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$
if k = s-1
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$
So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0
I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.
statistics proof-verification probability-distributions
asked Dec 30 '18 at 9:26
dsamaradsamara
12
$endgroup$
$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03
add a comment |
$begingroup$
It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:
$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$
= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$
= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$
if k = s-1
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$
So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0
I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.
statistics proof-verification probability-distributions
asked Dec 30 '18 at 9:26
dsamaradsamara
12
$endgroup$
It isn't proved in my study book of bio-statistics but suggested as an exercise. I tried the following method:
$PMF(y) = a*((PMF(x))^b$
= $a*((k/θ)((x/θ)^(k-1))e^(-(x/θ)^k))^b$
= $a*(k/θ)^b((x/θ)^(kb-b))(e^(-(x/θ)^(kb)))$
= $a*(k^b/θ^b)(x^k/θ^k)(e^(-x^(k*b)/θ^(k*b)))$
if k = s-1
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-x^(bl-b)/θ^(bl-b)))$
= $a*(s-1^b/θ^b)(x^(s-1)/θ^(s-1))(e^(-(x/θ)^s))$
So if I didn't make a mistake, I become something that looks like a Weibull distribution, but $a(s-1^b/θ^b)$ should be s-1/θ. I have no idea how I can become that if a,b is random and a,b>0
I searched on the web, but I couldn't find something similar. I hope someone can help me or suggest an other approach.
statistics proof-verification probability-distributions
statistics proof-verification probability-distributions
asked Dec 30 '18 at 9:26
dsamaradsamara
12
asked Dec 30 '18 at 9:26
dsamaradsamara
12
edited Dec 30 '18 at 9:41
dsamara
asked Dec 30 '18 at 9:26
dsamaradsamara
12
asked Dec 30 '18 at 9:26
dsamaradsamara
12
asked Dec 30 '18 at 9:26
dsamaradsamara
12
12
$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03
add a comment |
$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03
$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03
$begingroup$
Here's a MathJax tutorial.
$endgroup$
– StubbornAtom
Dec 30 '18 at 10:03
add a comment |
1 Answer
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$begingroup$
Other approach:
Let $U$ have standard exponential distribution.
Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
parameters $theta$ and $k$.
Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.
Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
showing directly that $Y$ has Weibull distribution with parameters
$atheta^{b}$ and $kb^{-1}$.
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Other approach:
Let $U$ have standard exponential distribution.
Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
parameters $theta$ and $k$.
Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.
Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
showing directly that $Y$ has Weibull distribution with parameters
$atheta^{b}$ and $kb^{-1}$.
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Other approach:
Let $U$ have standard exponential distribution.
Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
parameters $theta$ and $k$.
Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.
Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
showing directly that $Y$ has Weibull distribution with parameters
$atheta^{b}$ and $kb^{-1}$.
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Other approach:
Let $U$ have standard exponential distribution.
Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
parameters $theta$ and $k$.
Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.
Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
showing directly that $Y$ has Weibull distribution with parameters
$atheta^{b}$ and $kb^{-1}$.
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
$endgroup$
Other approach:
Let $U$ have standard exponential distribution.
Then for $x>0$: $$Pleft(theta U^{frac{1}{k}}>xright)=Pleft(Ugeqleft(frac{x}{theta}right)^{k}right)=e^{-left(frac{x}{theta}right)^{k}}$$
showing that $theta U^{frac{1}{k}}$ has Weibull distribution with
parameters $theta$ and $k$.
Conversely if $W$ has Weibull distribution then $U:=left(frac{W}{theta}right)^{k}$
has standard exponential distribution and we can write $W=theta U^{frac{1}{k}}$.
Then $Y=aX^{b}=aleft(theta U^{frac{1}{k}}right)^{b}=atheta^{b}U^{frac{1}{kb^{-1}}}$
showing directly that $Y$ has Weibull distribution with parameters
$atheta^{b}$ and $kb^{-1}$.
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
answered Dec 30 '18 at 9:55
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
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$begingroup$
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– StubbornAtom
Dec 30 '18 at 10:03