Mixing
Let X ~ Normal Standar find X given X > 0
$begingroup$
Let $X ~ N (0,1)$
Find the conditional PDF of $X$ given $X> 0$
I understand that this would be given by $ frac{f(x,x>0)}{f(x>0)}$, but I really do not know how to start, if you can give me ideas I appreciate.
probability statistics probability-distributions
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
$endgroup$
add a comment |
$begingroup$
Let $X ~ N (0,1)$
Find the conditional PDF of $X$ given $X> 0$
I understand that this would be given by $ frac{f(x,x>0)}{f(x>0)}$, but I really do not know how to start, if you can give me ideas I appreciate.
probability statistics probability-distributions
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
$endgroup$
2
$begingroup$
The expression $f(x>0)$ is nonsense! Domains and definitions are important: $f$ is a PDF, so it is a function of a single variable $x$, it is not defined for events like ${X>0}$. What you want to find is first the conditional distribution $$mathbb{P}(Xleq x | X>0)=frac{mathbb{P}(0<Xleq x)}{mathbb{P}(X>0)}$$ and then either identify the density as the answer below does or differentiate to finish.
$endgroup$
– LoveTooNap29
Jan 12 at 1:33
add a comment |
$begingroup$
Let $X ~ N (0,1)$
Find the conditional PDF of $X$ given $X> 0$
I understand that this would be given by $ frac{f(x,x>0)}{f(x>0)}$, but I really do not know how to start, if you can give me ideas I appreciate.
probability statistics probability-distributions
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
$endgroup$
Let $X ~ N (0,1)$
Find the conditional PDF of $X$ given $X> 0$
I understand that this would be given by $ frac{f(x,x>0)}{f(x>0)}$, but I really do not know how to start, if you can give me ideas I appreciate.
probability statistics probability-distributions
probability statistics probability-distributions
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
asked Jan 11 at 21:22
Cristian PerdomoCristian Perdomo
61
61
2
$begingroup$
The expression $f(x>0)$ is nonsense! Domains and definitions are important: $f$ is a PDF, so it is a function of a single variable $x$, it is not defined for events like ${X>0}$. What you want to find is first the conditional distribution $$mathbb{P}(Xleq x | X>0)=frac{mathbb{P}(0<Xleq x)}{mathbb{P}(X>0)}$$ and then either identify the density as the answer below does or differentiate to finish.
$endgroup$
– LoveTooNap29
Jan 12 at 1:33
add a comment |
2
$begingroup$
The expression $f(x>0)$ is nonsense! Domains and definitions are important: $f$ is a PDF, so it is a function of a single variable $x$, it is not defined for events like ${X>0}$. What you want to find is first the conditional distribution $$mathbb{P}(Xleq x | X>0)=frac{mathbb{P}(0<Xleq x)}{mathbb{P}(X>0)}$$ and then either identify the density as the answer below does or differentiate to finish.
$endgroup$
– LoveTooNap29
Jan 12 at 1:33
2
2
$begingroup$
The expression $f(x>0)$ is nonsense! Domains and definitions are important: $f$ is a PDF, so it is a function of a single variable $x$, it is not defined for events like ${X>0}$. What you want to find is first the conditional distribution $$mathbb{P}(Xleq x | X>0)=frac{mathbb{P}(0<Xleq x)}{mathbb{P}(X>0)}$$ and then either identify the density as the answer below does or differentiate to finish.
$endgroup$
– LoveTooNap29
Jan 12 at 1:33
$begingroup$
The expression $f(x>0)$ is nonsense! Domains and definitions are important: $f$ is a PDF, so it is a function of a single variable $x$, it is not defined for events like ${X>0}$. What you want to find is first the conditional distribution $$mathbb{P}(Xleq x | X>0)=frac{mathbb{P}(0<Xleq x)}{mathbb{P}(X>0)}$$ and then either identify the density as the answer below does or differentiate to finish.
$endgroup$
– LoveTooNap29
Jan 12 at 1:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $Y$ denote the R.V. $X|X>0$. By definition, the support of $Y$ is $[0, infty)$.
To compute the PDF, let's compute the CDF first.
$$P(Y>y) = P(X>y|X>0) = frac{P(X>y, X>0)}{P(X>0)}$$.
Since $y geq 0$, the numerator is simply $P(X>y)$. Since $X sim N(0,1)$, the denominator is ${1 over 2}$. Thus,
$$P(Y>y) = 2P(X>y)$$
Now you can see that $f_Y(y) = sqrt{{2 over pi}}e^{-y^2/2}$ for $y geq 0$ and $0$ otherwise.
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
$endgroup$
add a comment |
$begingroup$
Let $g$ be the density of $X$. ($g(x)=frac 1 {sqrt {2pi}} e^{-x^{2}/2})$. $P{Xleq x |X>0}=frac {int_0^{x} g(t)dt} {int_0^{infty} g(t)dt}=2int_0^{x} g(t)dt$ for $x>0$ and $0$ for $x<0$. Hence the density function is $2g(x)$for $x>0$, $0$ for $x<0$.
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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oldest
votes
$begingroup$
Let $Y$ denote the R.V. $X|X>0$. By definition, the support of $Y$ is $[0, infty)$.
To compute the PDF, let's compute the CDF first.
$$P(Y>y) = P(X>y|X>0) = frac{P(X>y, X>0)}{P(X>0)}$$.
Since $y geq 0$, the numerator is simply $P(X>y)$. Since $X sim N(0,1)$, the denominator is ${1 over 2}$. Thus,
$$P(Y>y) = 2P(X>y)$$
Now you can see that $f_Y(y) = sqrt{{2 over pi}}e^{-y^2/2}$ for $y geq 0$ and $0$ otherwise.
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
$endgroup$
add a comment |
$begingroup$
Let $Y$ denote the R.V. $X|X>0$. By definition, the support of $Y$ is $[0, infty)$.
To compute the PDF, let's compute the CDF first.
$$P(Y>y) = P(X>y|X>0) = frac{P(X>y, X>0)}{P(X>0)}$$.
Since $y geq 0$, the numerator is simply $P(X>y)$. Since $X sim N(0,1)$, the denominator is ${1 over 2}$. Thus,
$$P(Y>y) = 2P(X>y)$$
Now you can see that $f_Y(y) = sqrt{{2 over pi}}e^{-y^2/2}$ for $y geq 0$ and $0$ otherwise.
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
$endgroup$
add a comment |
$begingroup$
Let $Y$ denote the R.V. $X|X>0$. By definition, the support of $Y$ is $[0, infty)$.
To compute the PDF, let's compute the CDF first.
$$P(Y>y) = P(X>y|X>0) = frac{P(X>y, X>0)}{P(X>0)}$$.
Since $y geq 0$, the numerator is simply $P(X>y)$. Since $X sim N(0,1)$, the denominator is ${1 over 2}$. Thus,
$$P(Y>y) = 2P(X>y)$$
Now you can see that $f_Y(y) = sqrt{{2 over pi}}e^{-y^2/2}$ for $y geq 0$ and $0$ otherwise.
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
$endgroup$
Let $Y$ denote the R.V. $X|X>0$. By definition, the support of $Y$ is $[0, infty)$.
To compute the PDF, let's compute the CDF first.
$$P(Y>y) = P(X>y|X>0) = frac{P(X>y, X>0)}{P(X>0)}$$.
Since $y geq 0$, the numerator is simply $P(X>y)$. Since $X sim N(0,1)$, the denominator is ${1 over 2}$. Thus,
$$P(Y>y) = 2P(X>y)$$
Now you can see that $f_Y(y) = sqrt{{2 over pi}}e^{-y^2/2}$ for $y geq 0$ and $0$ otherwise.
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
answered Jan 12 at 0:31
Aditya DuaAditya Dua
1,16918
1,16918
add a comment |
add a comment |
$begingroup$
Let $g$ be the density of $X$. ($g(x)=frac 1 {sqrt {2pi}} e^{-x^{2}/2})$. $P{Xleq x |X>0}=frac {int_0^{x} g(t)dt} {int_0^{infty} g(t)dt}=2int_0^{x} g(t)dt$ for $x>0$ and $0$ for $x<0$. Hence the density function is $2g(x)$for $x>0$, $0$ for $x<0$.
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Let $g$ be the density of $X$. ($g(x)=frac 1 {sqrt {2pi}} e^{-x^{2}/2})$. $P{Xleq x |X>0}=frac {int_0^{x} g(t)dt} {int_0^{infty} g(t)dt}=2int_0^{x} g(t)dt$ for $x>0$ and $0$ for $x<0$. Hence the density function is $2g(x)$for $x>0$, $0$ for $x<0$.
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
add a comment |
$begingroup$
Let $g$ be the density of $X$. ($g(x)=frac 1 {sqrt {2pi}} e^{-x^{2}/2})$. $P{Xleq x |X>0}=frac {int_0^{x} g(t)dt} {int_0^{infty} g(t)dt}=2int_0^{x} g(t)dt$ for $x>0$ and $0$ for $x<0$. Hence the density function is $2g(x)$for $x>0$, $0$ for $x<0$.
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
$endgroup$
Let $g$ be the density of $X$. ($g(x)=frac 1 {sqrt {2pi}} e^{-x^{2}/2})$. $P{Xleq x |X>0}=frac {int_0^{x} g(t)dt} {int_0^{infty} g(t)dt}=2int_0^{x} g(t)dt$ for $x>0$ and $0$ for $x<0$. Hence the density function is $2g(x)$for $x>0$, $0$ for $x<0$.
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
edited Jan 12 at 0:43
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
answered Jan 11 at 23:38
Kavi Rama MurthyKavi Rama Murthy
59k42161
59k42161
add a comment |
add a comment |
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$begingroup$
The expression $f(x>0)$ is nonsense! Domains and definitions are important: $f$ is a PDF, so it is a function of a single variable $x$, it is not defined for events like ${X>0}$. What you want to find is first the conditional distribution $$mathbb{P}(Xleq x | X>0)=frac{mathbb{P}(0<Xleq x)}{mathbb{P}(X>0)}$$ and then either identify the density as the answer below does or differentiate to finish.
$endgroup$
– LoveTooNap29
Jan 12 at 1:33