CDF and PDF of standard normal random variable
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Trying to find the CDF and PDF for $Y=2+|X|$ where $X$ is a standard normal random variable.
Now if am not mistaken then I need to use for the CDF
$int_{-infty}^Xfrac{1}{sqrt{2pi}}e^{-t^2/2},dt$
and the PDF is just basically the derivative of that in this case.
My problem is I am not sure how do I deal exactly with the absolute value here.
So, I just want to check if my steps are correct below?
My attempt for CDF:
$$P(Yleq t) = P(2+|X|leq t)=P(|X|leq t-2)$$
$$P(X leq t-2)-P(X leq 2-t)= Phi(X)=int_{2-t}^{t-2}frac{1}{sqrt{2pi}}e^{-frac{tau^2}{2}},dtau =$$
$$=frac{1}{sqrt{2pi}}(-2e^frac{2-t}{2}+2e^frac{t-2}{2})$$
probability probability-distributions
asked Nov 13 at 20:09
JKM
4714
add a comment |
up vote
1
down vote
favorite
Trying to find the CDF and PDF for $Y=2+|X|$ where $X$ is a standard normal random variable.
Now if am not mistaken then I need to use for the CDF
$int_{-infty}^Xfrac{1}{sqrt{2pi}}e^{-t^2/2},dt$
and the PDF is just basically the derivative of that in this case.
My problem is I am not sure how do I deal exactly with the absolute value here.
So, I just want to check if my steps are correct below?
My attempt for CDF:
$$P(Yleq t) = P(2+|X|leq t)=P(|X|leq t-2)$$
$$P(X leq t-2)-P(X leq 2-t)= Phi(X)=int_{2-t}^{t-2}frac{1}{sqrt{2pi}}e^{-frac{tau^2}{2}},dtau =$$
$$=frac{1}{sqrt{2pi}}(-2e^frac{2-t}{2}+2e^frac{t-2}{2})$$
probability probability-distributions
asked Nov 13 at 20:09
JKM
4714
All looks good except the last step
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 20:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Trying to find the CDF and PDF for $Y=2+|X|$ where $X$ is a standard normal random variable.
Now if am not mistaken then I need to use for the CDF
$int_{-infty}^Xfrac{1}{sqrt{2pi}}e^{-t^2/2},dt$
and the PDF is just basically the derivative of that in this case.
My problem is I am not sure how do I deal exactly with the absolute value here.
So, I just want to check if my steps are correct below?
My attempt for CDF:
$$P(Yleq t) = P(2+|X|leq t)=P(|X|leq t-2)$$
$$P(X leq t-2)-P(X leq 2-t)= Phi(X)=int_{2-t}^{t-2}frac{1}{sqrt{2pi}}e^{-frac{tau^2}{2}},dtau =$$
$$=frac{1}{sqrt{2pi}}(-2e^frac{2-t}{2}+2e^frac{t-2}{2})$$
probability probability-distributions
asked Nov 13 at 20:09
JKM
4714
Trying to find the CDF and PDF for $Y=2+|X|$ where $X$ is a standard normal random variable.
Now if am not mistaken then I need to use for the CDF
$int_{-infty}^Xfrac{1}{sqrt{2pi}}e^{-t^2/2},dt$
and the PDF is just basically the derivative of that in this case.
My problem is I am not sure how do I deal exactly with the absolute value here.
So, I just want to check if my steps are correct below?
My attempt for CDF:
$$P(Yleq t) = P(2+|X|leq t)=P(|X|leq t-2)$$
$$P(X leq t-2)-P(X leq 2-t)= Phi(X)=int_{2-t}^{t-2}frac{1}{sqrt{2pi}}e^{-frac{tau^2}{2}},dtau =$$
$$=frac{1}{sqrt{2pi}}(-2e^frac{2-t}{2}+2e^frac{t-2}{2})$$
probability probability-distributions
probability probability-distributions
asked Nov 13 at 20:09
JKM
4714
asked Nov 13 at 20:09
JKM
4714
asked Nov 13 at 20:09
JKM
4714
asked Nov 13 at 20:09
JKM
4714
asked Nov 13 at 20:09
JKM
4714
4714
All looks good except the last step
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 20:30
add a comment |
All looks good except the last step
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 20:30
All looks good except the last step
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 20:30
All looks good except the last step
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 20:30
add a comment |
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All looks good except the last step
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 20:30