What does this distribution mean?

c2q

New Member
#1
When it says:

the marginal distribution of Y is given by

1/Y ~ Exponential(lamda)

does that mean the distribution is written has:

1/(lamda)e^(-lamda)(y)

??

Edit: Nevermind i think my first thought process is wrong, i believe its:

(lamda)e^(-lamda)(1/y)

if im right.
 
Last edited:

BGM

TS Contributor
#2
if the probability density function of [math] X = \frac {1} {Y} [/math] is

[math] f_X(x;\lambda) = \lambda e^{-\lambda x} , x > 0, \lambda > 0 [/math]

Then it is easy to show that the probability density function of [math] Y [/math] is

[math] f_Y(y;\lambda) = \lambda \frac {1} {y^2} e^{-\lambda \frac {1} {y}},
y > 0, \lambda > 0 [/math]

Note: The expected value of [math] Y [/math] is may not exist / finite.
 

c2q

New Member
#3
if the probability density function of [math] X = \frac {1} {Y} [/math] is

[math] f_X(x;\lambda) = \lambda e^{-\lambda x} , x > 0, \lambda > 0 [/math]

Then it is easy to show that the probability density function of [math] Y [/math] is

[math] f_Y(y;\lambda) = \lambda \frac {1} {y^2} e^{-\lambda \frac {1} {y}},
y > 0, \lambda > 0 [/math]

Note: The expected value of [math] Y [/math] is may not exist / finite.
just wondering, how did you get the 1/y^2 in the second equation of the PDF of Y? I thought its basically like that without the 1/y^2 if you can explain that.
 

BGM

TS Contributor
#4
Basically you can try the CDF approach or the Jacobian Transformation on pdf
with 1-to-1 transformation.

1. Consider

[math] F_Y(y) = \Pr\{Y \leq y\}
= \Pr\left\{\frac {1} {Y} \geq \frac {1} {y} \right\}
= 1 - \Pr\left\{X \leq \frac {1} {y} \right\}
= 1 - F_X\left(\frac {1} {y} \right) [/math]

Since we can get the pdf by differentiating the CDF, we have

[math] f_Y(y) = \frac {dF_Y(y)} {dy}
= \frac {d} {dy} [ 1 - F_X(y^{-1}) ]
= - \frac {d} {dy^{-1}} F_X(y^{-1}) \frac {dy^{-1}} {dy}
= \frac {1} {y^2} f_X(y^{-1})
= \frac {1} {y^2} \lambda e^{-\lambda \frac {1} {y}} [/math]

2. Since [math] y = \frac {1} {x} \iff x = \frac {1} {y} [/math]

we have the absolute value of the Jacobian determinant

[math] |J| = \left|\frac {dx} {dy}\right|
= \left|-\frac {1} {y^2} \right| = \frac {1} {y^2} [/math]

and thus

[math] f_Y(y) = |J|f_X(x) = \frac {1} {y^2} f_X(y^{-1})
= \frac {1} {y^2} \lambda e^{-\lambda \frac {1} {y}} [/math]