Let X and Y iid be ~ N(0,1)[Gaussian]
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Given Z=XY.Find the conditional pdf Z given X=x.
The way I calculated assumed W=X, then proceeded with Jacobian and replaced W=X in the final answer.
Not sure whether it is right or not.
Answer i got is $ frac{1}{sqrt{2pi}}e^-{frac{z^2}{2x^2}} $.
Could anyone help me if the procedure/answer is wrong
probability probability-theory probability-distributions random-variables conditional-probability
asked Nov 15 at 20:45
Pramod_achar
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up vote
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down vote
favorite
Given Z=XY.Find the conditional pdf Z given X=x.
The way I calculated assumed W=X, then proceeded with Jacobian and replaced W=X in the final answer.
Not sure whether it is right or not.
Answer i got is $ frac{1}{sqrt{2pi}}e^-{frac{z^2}{2x^2}} $.
Could anyone help me if the procedure/answer is wrong
probability probability-theory probability-distributions random-variables conditional-probability
asked Nov 15 at 20:45
Pramod_achar
13
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given Z=XY.Find the conditional pdf Z given X=x.
The way I calculated assumed W=X, then proceeded with Jacobian and replaced W=X in the final answer.
Not sure whether it is right or not.
Answer i got is $ frac{1}{sqrt{2pi}}e^-{frac{z^2}{2x^2}} $.
Could anyone help me if the procedure/answer is wrong
probability probability-theory probability-distributions random-variables conditional-probability
asked Nov 15 at 20:45
Pramod_achar
13
Given Z=XY.Find the conditional pdf Z given X=x.
The way I calculated assumed W=X, then proceeded with Jacobian and replaced W=X in the final answer.
Not sure whether it is right or not.
Answer i got is $ frac{1}{sqrt{2pi}}e^-{frac{z^2}{2x^2}} $.
Could anyone help me if the procedure/answer is wrong
probability probability-theory probability-distributions random-variables conditional-probability
probability probability-theory probability-distributions random-variables conditional-probability
asked Nov 15 at 20:45
Pramod_achar
13
asked Nov 15 at 20:45
Pramod_achar
13
asked Nov 15 at 20:45
Pramod_achar
13
asked Nov 15 at 20:45
Pramod_achar
13
asked Nov 15 at 20:45
Pramod_achar
13
13
add a comment |
add a comment |
1 Answer
1
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Not quite; your result doesn't integrate to $1$. For $x>0$,$$P(Zle z|Xle x)=Pbigg(Ylefrac{z}{x}bigg)=Phibigg(frac{z}{x}bigg),$$giving a pdf for $Z$ of $$frac{1}{x}phibigg(frac{z}{x}bigg)=frac{1}{xsqrt{2pi}}exp-frac{z^2}{2x^2}.$$You can repeat this for $x<0$. In general, the PDF of $z$ is $$frac{1}{|x|sqrt{2pi}}exp-frac{z^2}{2x^2}.$$
answered Nov 15 at 22:44
J.G.
19.6k21932
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Not quite; your result doesn't integrate to $1$. For $x>0$,$$P(Zle z|Xle x)=Pbigg(Ylefrac{z}{x}bigg)=Phibigg(frac{z}{x}bigg),$$giving a pdf for $Z$ of $$frac{1}{x}phibigg(frac{z}{x}bigg)=frac{1}{xsqrt{2pi}}exp-frac{z^2}{2x^2}.$$You can repeat this for $x<0$. In general, the PDF of $z$ is $$frac{1}{|x|sqrt{2pi}}exp-frac{z^2}{2x^2}.$$
answered Nov 15 at 22:44
J.G.
19.6k21932
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
add a comment |
up vote
0
down vote
Not quite; your result doesn't integrate to $1$. For $x>0$,$$P(Zle z|Xle x)=Pbigg(Ylefrac{z}{x}bigg)=Phibigg(frac{z}{x}bigg),$$giving a pdf for $Z$ of $$frac{1}{x}phibigg(frac{z}{x}bigg)=frac{1}{xsqrt{2pi}}exp-frac{z^2}{2x^2}.$$You can repeat this for $x<0$. In general, the PDF of $z$ is $$frac{1}{|x|sqrt{2pi}}exp-frac{z^2}{2x^2}.$$
answered Nov 15 at 22:44
J.G.
19.6k21932
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
add a comment |
up vote
0
down vote
up vote
0
down vote
Not quite; your result doesn't integrate to $1$. For $x>0$,$$P(Zle z|Xle x)=Pbigg(Ylefrac{z}{x}bigg)=Phibigg(frac{z}{x}bigg),$$giving a pdf for $Z$ of $$frac{1}{x}phibigg(frac{z}{x}bigg)=frac{1}{xsqrt{2pi}}exp-frac{z^2}{2x^2}.$$You can repeat this for $x<0$. In general, the PDF of $z$ is $$frac{1}{|x|sqrt{2pi}}exp-frac{z^2}{2x^2}.$$
answered Nov 15 at 22:44
J.G.
19.6k21932
Not quite; your result doesn't integrate to $1$. For $x>0$,$$P(Zle z|Xle x)=Pbigg(Ylefrac{z}{x}bigg)=Phibigg(frac{z}{x}bigg),$$giving a pdf for $Z$ of $$frac{1}{x}phibigg(frac{z}{x}bigg)=frac{1}{xsqrt{2pi}}exp-frac{z^2}{2x^2}.$$You can repeat this for $x<0$. In general, the PDF of $z$ is $$frac{1}{|x|sqrt{2pi}}exp-frac{z^2}{2x^2}.$$
answered Nov 15 at 22:44
J.G.
19.6k21932
answered Nov 15 at 22:44
J.G.
19.6k21932
answered Nov 15 at 22:44
J.G.
19.6k21932
answered Nov 15 at 22:44
J.G.
19.6k21932
19.6k21932
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
add a comment |
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
thank you, How do I learn these shortcut methods. Any material/book do you suggest
– Pramod_achar
Nov 16 at 4:17
add a comment |
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