Question

Suppose that the p.d.f. of X is as follows:

\(\begin{aligned}f\left( x \right) &= e{}^{ - x},x > 0\\ &= 0,x \le 0\end{aligned}\)

Determine the p.d.f. of \({\bf{Y = }}{{\bf{X}}^{\frac{{\bf{1}}}{{\bf{2}}}}}\)

Short answer

The PDF of \(Y = {X^{\frac{1}{2}}}:2y{e^{ - {y^2}}},y > 0\)

Step-by-step solution

Given information

The random variable X has an exponential distribution with parameter 1\(X \sim \exp \left( 1 \right)\).

Obtain the PDF and CDF of X

The pdf of an exponential distribution is obtained by using the formula: \({e^{ - \lambda x}},x > 0\)

Here \(\lambda = 1\)

The pdf of X is -

\(\begin{array}{c}{f_x} = e{}^{ - x},x > 0\\ = 0,x \le 0\end{array}\)

The CDF of an exponential distribution is obtained by using the formula:

The CDF of X is defined as:

\(\begin{aligned}{F_X}\left( x \right) &= P\left( {X \le x} \right)\\ &= \int\limits_0^x {{e^{ - x}}} dx\\ &= 1 - {e^{ - x}},x > 0\end{aligned}\)

Create a new variable, Y, and use the CDF approach

The new variable is defined as \(Y = {X^{\frac{1}{2}}}\)

A CDF approach is a method of random variable transformation wherein the pdf of the new variable is fetched from the CDF of the new variable, which is in terms of the CDF of the old variable.

The CDF approach steps,

• We substitute the Y variable in the CDF formula in a CDF approach.
• We then substitute Y in terms of X.
• We reduce this form until we bring the CDF in terms of X.
• Since we have already calculated the CDF of X, we replace the form of variable y in the formula for X.
• In the final step, we get the CDF of the Y variable as an expression of the CDF of X with y variables.

By using the CDF approach.

\(\begin{aligned}{F_{Y = }}\left( y \right) &= P\left( {Y \le y} \right)\\ &= P\left( {{X^{\frac{1}{2}}} \le y} \right)\\ &= P\left( {X \le {y^2}} \right)\\ &= {F_X}\left( {{y^2}} \right)\\ &= 1 - {e^{ - {y^2}}}\end{aligned}\)

Therefore, the CDF of Y is \(1 - {e^{ - {y^2}}}\)

Convert the CDF into PDF

The pdf is obtained from CDF by differentiating it with respect to the variable.

\(\begin{aligned}{f_y} &= \frac{d}{{dx}}\left( {{F_Y}\left( y \right)} \right)\\ &= \frac{d}{{dx}}\left( {1 - {e^{ - {y^2}}}} \right)\\ &= 2y{e^{ - {y^2}}},y > 0\end{aligned}\)

Therefore, the pdf of the variable Y is \(2y{e^{ - {y^2}}},y > 0\)