Median of Rayleigh Distribution
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I am not sure how to solve the following problem:
The probability density function of the Rayleigh distribution is,
$ f(x;α) = fracxα^2 e^frac-x^22α^2, x ≥ 0, $
where α is the scale parameter of the distribution. Find the median of the Rayleigh distribution.
I need to derive the median of the distribution, but do not know how to do so. Thoughts?
Thanks!
pdf median rayleigh
asked 1 hour ago
Joe Ademo
254
add a comment |Â
up vote
1
down vote
favorite
I am not sure how to solve the following problem:
The probability density function of the Rayleigh distribution is,
$ f(x;α) = fracxα^2 e^frac-x^22α^2, x ≥ 0, $
where α is the scale parameter of the distribution. Find the median of the Rayleigh distribution.
I need to derive the median of the distribution, but do not know how to do so. Thoughts?
Thanks!
pdf median rayleigh
asked 1 hour ago
Joe Ademo
254
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am not sure how to solve the following problem:
The probability density function of the Rayleigh distribution is,
$ f(x;α) = fracxα^2 e^frac-x^22α^2, x ≥ 0, $
where α is the scale parameter of the distribution. Find the median of the Rayleigh distribution.
I need to derive the median of the distribution, but do not know how to do so. Thoughts?
Thanks!
pdf median rayleigh
asked 1 hour ago
Joe Ademo
254
I am not sure how to solve the following problem:
The probability density function of the Rayleigh distribution is,
$ f(x;α) = fracxα^2 e^frac-x^22α^2, x ≥ 0, $
where α is the scale parameter of the distribution. Find the median of the Rayleigh distribution.
I need to derive the median of the distribution, but do not know how to do so. Thoughts?
Thanks!
pdf median rayleigh
pdf median rayleigh
asked 1 hour ago
Joe Ademo
254
asked 1 hour ago
Joe Ademo
254
edited 1 hour ago
asked 1 hour ago
Joe Ademo
254
asked 1 hour ago
Joe Ademo
254
asked 1 hour ago
Joe Ademo
254
254
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The Rayleigh distribution has cumulative distribution function (CDF) $F_X(x) = 1-texte^frac-x^22alpha^2$.
Denote the median $q_50$.
Starting with the CDF...
$$beginalign
1-texte^frac-q_50^22alpha^2 &= 0.5 \
texte^frac-q_50^22alpha^2 &= 0.5 \
frac-q_50^22alpha^2 &= textln(0.5) \
-q_50^2 &= 2alpha^2 textln(0.5) \
\
q_50 &=alpha sqrt-2 textln(0.5) \
&= alpha sqrt2textln(2) quad quad square
endalign$$
answered 1 hour ago
SecretAgentMan
610121
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The Rayleigh distribution has cumulative distribution function (CDF) $F_X(x) = 1-texte^frac-x^22alpha^2$.
Denote the median $q_50$.
Starting with the CDF...
$$beginalign
1-texte^frac-q_50^22alpha^2 &= 0.5 \
texte^frac-q_50^22alpha^2 &= 0.5 \
frac-q_50^22alpha^2 &= textln(0.5) \
-q_50^2 &= 2alpha^2 textln(0.5) \
\
q_50 &=alpha sqrt-2 textln(0.5) \
&= alpha sqrt2textln(2) quad quad square
endalign$$
answered 1 hour ago
SecretAgentMan
610121
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
add a comment |Â
up vote
3
down vote
accepted
The Rayleigh distribution has cumulative distribution function (CDF) $F_X(x) = 1-texte^frac-x^22alpha^2$.
Denote the median $q_50$.
Starting with the CDF...
$$beginalign
1-texte^frac-q_50^22alpha^2 &= 0.5 \
texte^frac-q_50^22alpha^2 &= 0.5 \
frac-q_50^22alpha^2 &= textln(0.5) \
-q_50^2 &= 2alpha^2 textln(0.5) \
\
q_50 &=alpha sqrt-2 textln(0.5) \
&= alpha sqrt2textln(2) quad quad square
endalign$$
answered 1 hour ago
SecretAgentMan
610121
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The Rayleigh distribution has cumulative distribution function (CDF) $F_X(x) = 1-texte^frac-x^22alpha^2$.
Denote the median $q_50$.
Starting with the CDF...
$$beginalign
1-texte^frac-q_50^22alpha^2 &= 0.5 \
texte^frac-q_50^22alpha^2 &= 0.5 \
frac-q_50^22alpha^2 &= textln(0.5) \
-q_50^2 &= 2alpha^2 textln(0.5) \
\
q_50 &=alpha sqrt-2 textln(0.5) \
&= alpha sqrt2textln(2) quad quad square
endalign$$
answered 1 hour ago
SecretAgentMan
610121
The Rayleigh distribution has cumulative distribution function (CDF) $F_X(x) = 1-texte^frac-x^22alpha^2$.
Denote the median $q_50$.
Starting with the CDF...
$$beginalign
1-texte^frac-q_50^22alpha^2 &= 0.5 \
texte^frac-q_50^22alpha^2 &= 0.5 \
frac-q_50^22alpha^2 &= textln(0.5) \
-q_50^2 &= 2alpha^2 textln(0.5) \
\
q_50 &=alpha sqrt-2 textln(0.5) \
&= alpha sqrt2textln(2) quad quad square
endalign$$
answered 1 hour ago
SecretAgentMan
610121
answered 1 hour ago
SecretAgentMan
610121
answered 1 hour ago
SecretAgentMan
610121
answered 1 hour ago
SecretAgentMan
610121
610121
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
add a comment |Â
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Right, because this is a CRV I should use the CDF. Thanks!!
– Joe Ademo
55 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least.
– SecretAgentMan
53 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help.
– Joe Ademo
49 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
You could use the density, especially if you love integration by parts... Solve the equation $int_0^q_50 f_X(x)dx = 0.5$.
– SecretAgentMan
37 mins ago
add a comment |Â
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