Mixing
Compute the density, distribution function
Compute the density function of Z = X2/X1 given that X2 and X1 both have the normal distribution with mean 0, variance 1 as their density.
I got this far:
$$
F(z) = P( X_2 leq X_1 z) = intint frac{1}{2 pi} e^{-x^2-y^2} dx dy
$$
change of variables
$$
int^{+infty}_0 int^{tan^{-1}(z + 2pi)}_{tan^{-1}( z + pi)} frac{1}{2pi}e^{-r^2}r dr dtheta = 1/2
$$
Now the textbook answer is $tan^{-1} z / pi$ without any explanations and I have trouble understanding it as I am trying to self learn probability. Any help greatly appreciated.
probability density-function
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
add a comment |
Compute the density function of Z = X2/X1 given that X2 and X1 both have the normal distribution with mean 0, variance 1 as their density.
I got this far:
$$
F(z) = P( X_2 leq X_1 z) = intint frac{1}{2 pi} e^{-x^2-y^2} dx dy
$$
change of variables
$$
int^{+infty}_0 int^{tan^{-1}(z + 2pi)}_{tan^{-1}( z + pi)} frac{1}{2pi}e^{-r^2}r dr dtheta = 1/2
$$
Now the textbook answer is $tan^{-1} z / pi$ without any explanations and I have trouble understanding it as I am trying to self learn probability. Any help greatly appreciated.
probability density-function
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it?
– Will M.
Nov 21 '18 at 18:51
I thought the goal is to find an expression for $F(Z leq z)$ which is equivalent to $F(frac{X_2}{X_1} leq z)$ equivalent to $F(X_2 leq z X_1)$. z is a real number. edit because I screwed up the math typesetting
– Chen Ee Woon
Nov 21 '18 at 18:54
You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: math.stackexchange.com/questions/77873/….
– StubbornAtom
Nov 21 '18 at 18:58
Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it.
– Chen Ee Woon
Nov 21 '18 at 19:15
add a comment |
Compute the density function of Z = X2/X1 given that X2 and X1 both have the normal distribution with mean 0, variance 1 as their density.
I got this far:
$$
F(z) = P( X_2 leq X_1 z) = intint frac{1}{2 pi} e^{-x^2-y^2} dx dy
$$
change of variables
$$
int^{+infty}_0 int^{tan^{-1}(z + 2pi)}_{tan^{-1}( z + pi)} frac{1}{2pi}e^{-r^2}r dr dtheta = 1/2
$$
Now the textbook answer is $tan^{-1} z / pi$ without any explanations and I have trouble understanding it as I am trying to self learn probability. Any help greatly appreciated.
probability density-function
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
Compute the density function of Z = X2/X1 given that X2 and X1 both have the normal distribution with mean 0, variance 1 as their density.
I got this far:
$$
F(z) = P( X_2 leq X_1 z) = intint frac{1}{2 pi} e^{-x^2-y^2} dx dy
$$
change of variables
$$
int^{+infty}_0 int^{tan^{-1}(z + 2pi)}_{tan^{-1}( z + pi)} frac{1}{2pi}e^{-r^2}r dr dtheta = 1/2
$$
Now the textbook answer is $tan^{-1} z / pi$ without any explanations and I have trouble understanding it as I am trying to self learn probability. Any help greatly appreciated.
probability density-function
probability density-function
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
edited Nov 22 '18 at 5:04
Flying Dogfish
1808
1808
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
asked Nov 21 '18 at 18:46
Chen Ee WoonChen Ee Woon
6
6
On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it?
– Will M.
Nov 21 '18 at 18:51
I thought the goal is to find an expression for $F(Z leq z)$ which is equivalent to $F(frac{X_2}{X_1} leq z)$ equivalent to $F(X_2 leq z X_1)$. z is a real number. edit because I screwed up the math typesetting
– Chen Ee Woon
Nov 21 '18 at 18:54
You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: math.stackexchange.com/questions/77873/….
– StubbornAtom
Nov 21 '18 at 18:58
Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it.
– Chen Ee Woon
Nov 21 '18 at 19:15
add a comment |
On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it?
– Will M.
Nov 21 '18 at 18:51
I thought the goal is to find an expression for $F(Z leq z)$ which is equivalent to $F(frac{X_2}{X_1} leq z)$ equivalent to $F(X_2 leq z X_1)$. z is a real number. edit because I screwed up the math typesetting
– Chen Ee Woon
Nov 21 '18 at 18:54
You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: math.stackexchange.com/questions/77873/….
– StubbornAtom
Nov 21 '18 at 18:58
Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it.
– Chen Ee Woon
Nov 21 '18 at 19:15
On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it?
– Will M.
Nov 21 '18 at 18:51
On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it?
– Will M.
Nov 21 '18 at 18:51
I thought the goal is to find an expression for $F(Z leq z)$ which is equivalent to $F(frac{X_2}{X_1} leq z)$ equivalent to $F(X_2 leq z X_1)$. z is a real number. edit because I screwed up the math typesetting
– Chen Ee Woon
Nov 21 '18 at 18:54
I thought the goal is to find an expression for $F(Z leq z)$ which is equivalent to $F(frac{X_2}{X_1} leq z)$ equivalent to $F(X_2 leq z X_1)$. z is a real number. edit because I screwed up the math typesetting
– Chen Ee Woon
Nov 21 '18 at 18:54
You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: math.stackexchange.com/questions/77873/….
– StubbornAtom
Nov 21 '18 at 18:58
You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: math.stackexchange.com/questions/77873/….
– StubbornAtom
Nov 21 '18 at 18:58
Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it.
– Chen Ee Woon
Nov 21 '18 at 19:15
Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it.
– Chen Ee Woon
Nov 21 '18 at 19:15
add a comment |
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On the one hand, whomever voted down the question should provide an explanation as to why. On the other, your second equality, how did you get it?
– Will M.
Nov 21 '18 at 18:51
I thought the goal is to find an expression for $F(Z leq z)$ which is equivalent to $F(frac{X_2}{X_1} leq z)$ equivalent to $F(X_2 leq z X_1)$. z is a real number. edit because I screwed up the math typesetting
– Chen Ee Woon
Nov 21 '18 at 18:54
You are probably given that $X_1$ is independent of $X_2$. Then one of the many posts here that has an answer to this question: math.stackexchange.com/questions/77873/….
– StubbornAtom
Nov 21 '18 at 18:58
Many thanks. I made a blunder multiplying by a negative number without changing the inequality. And I didn't realise Will is hinting at it.
– Chen Ee Woon
Nov 21 '18 at 19:15