A random variable in a denominator
$begingroup$
For some random variable $X$ & $Y$, I need to calculate the pdf of $Z=frac{X}{Y}$. I've managed to calculate $$F_Z(t)=mathbb{P}left(frac{X}{Y}leq tright)=mathbb{P}(Xleq tcdot Y),,$$ but I didn't treat to the necessity of $Yne 0$. In what way in the answer should I treat this necessity?
probability probability-distributions random-variables fractions random-functions
edited Dec 14 '18 at 21:58
Batominovski
33k33293
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
$endgroup$
add a comment |
$begingroup$
For some random variable $X$ & $Y$, I need to calculate the pdf of $Z=frac{X}{Y}$. I've managed to calculate $$F_Z(t)=mathbb{P}left(frac{X}{Y}leq tright)=mathbb{P}(Xleq tcdot Y),,$$ but I didn't treat to the necessity of $Yne 0$. In what way in the answer should I treat this necessity?
probability probability-distributions random-variables fractions random-functions
edited Dec 14 '18 at 21:58
Batominovski
33k33293
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
$endgroup$
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
add a comment |
$begingroup$
For some random variable $X$ & $Y$, I need to calculate the pdf of $Z=frac{X}{Y}$. I've managed to calculate $$F_Z(t)=mathbb{P}left(frac{X}{Y}leq tright)=mathbb{P}(Xleq tcdot Y),,$$ but I didn't treat to the necessity of $Yne 0$. In what way in the answer should I treat this necessity?
probability probability-distributions random-variables fractions random-functions
edited Dec 14 '18 at 21:58
Batominovski
33k33293
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
$endgroup$
For some random variable $X$ & $Y$, I need to calculate the pdf of $Z=frac{X}{Y}$. I've managed to calculate $$F_Z(t)=mathbb{P}left(frac{X}{Y}leq tright)=mathbb{P}(Xleq tcdot Y),,$$ but I didn't treat to the necessity of $Yne 0$. In what way in the answer should I treat this necessity?
probability probability-distributions random-variables fractions random-functions
probability probability-distributions random-variables fractions random-functions
edited Dec 14 '18 at 21:58
Batominovski
33k33293
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
edited Dec 14 '18 at 21:58
Batominovski
33k33293
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
edited Dec 14 '18 at 21:58
Batominovski
33k33293
edited Dec 14 '18 at 21:58
Batominovski
33k33293
edited Dec 14 '18 at 21:58
Batominovski
33k33293
33k33293
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
asked Dec 14 '18 at 19:42
J. DoeJ. Doe
14510
14510
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
add a comment |
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
1
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
1
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
add a comment |
1 Answer
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$begingroup$
If $P(Y = 0) > 0$, then there is a nonzero probability that $Z$ is undefined due to dividing by zero.
However, if $P(Y = 0) = 0$ (for example, if $Y$ has a PDF), then the probability of $Z$ being undefined is zero, so you can essentially ignore that case.
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
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add a comment |
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1 Answer
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$begingroup$
If $P(Y = 0) > 0$, then there is a nonzero probability that $Z$ is undefined due to dividing by zero.
However, if $P(Y = 0) = 0$ (for example, if $Y$ has a PDF), then the probability of $Z$ being undefined is zero, so you can essentially ignore that case.
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
$endgroup$
add a comment |
$begingroup$
If $P(Y = 0) > 0$, then there is a nonzero probability that $Z$ is undefined due to dividing by zero.
However, if $P(Y = 0) = 0$ (for example, if $Y$ has a PDF), then the probability of $Z$ being undefined is zero, so you can essentially ignore that case.
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
$endgroup$
add a comment |
$begingroup$
If $P(Y = 0) > 0$, then there is a nonzero probability that $Z$ is undefined due to dividing by zero.
However, if $P(Y = 0) = 0$ (for example, if $Y$ has a PDF), then the probability of $Z$ being undefined is zero, so you can essentially ignore that case.
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
$endgroup$
If $P(Y = 0) > 0$, then there is a nonzero probability that $Z$ is undefined due to dividing by zero.
However, if $P(Y = 0) = 0$ (for example, if $Y$ has a PDF), then the probability of $Z$ being undefined is zero, so you can essentially ignore that case.
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
answered Dec 14 '18 at 20:13
angryavianangryavian
40.9k23380
40.9k23380
add a comment |
add a comment |
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$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16
1
$begingroup$
What you should show is $P(Y=0)=0$. This condition guarantees that $frac{X}{Y}$ is well-defined almost surely.
$endgroup$
– Song
Dec 14 '18 at 20:16