# What does this distribution mean?

#### c2q

##### New Member
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.

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#### BGM

##### TS Contributor
if the probability density function of $X = \frac {1} {Y}$ is

$f_X(x;\lambda) = \lambda e^{-\lambda x} , x > 0, \lambda > 0$

Then it is easy to show that the probability density function of $Y$ is

$f_Y(y;\lambda) = \lambda \frac {1} {y^2} e^{-\lambda \frac {1} {y}}, y > 0, \lambda > 0$

Note: The expected value of $Y$ is may not exist / finite.

#### c2q

##### New Member
if the probability density function of $X = \frac {1} {Y}$ is

$f_X(x;\lambda) = \lambda e^{-\lambda x} , x > 0, \lambda > 0$

Then it is easy to show that the probability density function of $Y$ is

$f_Y(y;\lambda) = \lambda \frac {1} {y^2} e^{-\lambda \frac {1} {y}}, y > 0, \lambda > 0$

Note: The expected value of $Y$ 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
Basically you can try the CDF approach or the Jacobian Transformation on pdf
with 1-to-1 transformation.

1. Consider

$F_Y(y) = \Pr\{Y \leq y\} = \Pr\left\{\frac {1} {Y} \geq \frac {1} {y} \right\} = 1 - \Pr\left\{X \leq \frac {1} {y} \right\} = 1 - F_X\left(\frac {1} {y} \right)$

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

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

2. Since $y = \frac {1} {x} \iff x = \frac {1} {y}$

we have the absolute value of the Jacobian determinant

$|J| = \left|\frac {dx} {dy}\right| = \left|-\frac {1} {y^2} \right| = \frac {1} {y^2}$

and thus

$f_Y(y) = |J|f_X(x) = \frac {1} {y^2} f_X(y^{-1}) = \frac {1} {y^2} \lambda e^{-\lambda \frac {1} {y}}$