Solving $P(2<X^2<5)$ given PDF $f(x)$ and CDF $F(x)$ of X
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.
For this case do I solve $2$ cases with
$P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$
My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.
probability
asked Aug 26 at 21:54
glockm15
1649
add a comment |Â
up vote
1
down vote
favorite
If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.
For this case do I solve $2$ cases with
$P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$
My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.
probability
asked Aug 26 at 21:54
glockm15
1649
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.
For this case do I solve $2$ cases with
$P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$
My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.
probability
asked Aug 26 at 21:54
glockm15
1649
If it was $P(2<X<5)$ I would just use $F(5)-F(2)$.
For this case do I solve $2$ cases with
$P(sqrt2<X<sqrt5)$ and $P(-sqrt 5 < X < - sqrt2)$
My random variable is defined for all real numbers so I think this is valid, is this correct? Note my pdf has an absolute value so the CDF is defined separately for $xle0$ and $x>0$ which also supports doing that.
probability
asked Aug 26 at 21:54
glockm15
1649
edited Aug 26 at 22:22
user550230
asked Aug 26 at 21:54
glockm15
1649
asked Aug 26 at 21:54
glockm15
1649
asked Aug 26 at 21:54
glockm15
1649
1649
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.
answered Aug 26 at 22:01
heropup
60.3k65895
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
1
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.
answered Aug 26 at 22:01
heropup
60.3k65895
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
1
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
add a comment |Â
up vote
1
down vote
accepted
You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.
answered Aug 26 at 22:01
heropup
60.3k65895
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
1
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.
answered Aug 26 at 22:01
heropup
60.3k65895
You have the right idea but it is important to note that $$-sqrt5 < -sqrt2,$$ thus your second probability expression should read $$Pr[-sqrt5 < X < -sqrt2],$$ not the other way around--which would give zero.
answered Aug 26 at 22:01
heropup
60.3k65895
answered Aug 26 at 22:01
heropup
60.3k65895
answered Aug 26 at 22:01
heropup
60.3k65895
answered Aug 26 at 22:01
heropup
60.3k65895
60.3k65895
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
1
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
add a comment |Â
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
1
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
So when I have found the probabilities for each case would it make more sense to add probability of the two answers.As x being positive or negative are independent events. Or just report each probability found for each case separately.
– glockm15
Aug 26 at 22:08
1
1
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
You do need to add the probabilities together; that is to say, $$Pr[2 < X^2 < 5] = Pr[-sqrt5 < X < -sqrt2] + Pr[sqrt2 < X < sqrt5].$$ Note that this works because the intervals over which you are evaluating those probabilities are disjoint.
– heropup
Aug 26 at 22:17
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2895569%2fsolving-p2x25-given-pdf-fx-and-cdf-fx-of-x%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
