Find the Density of Y = 1/X When X Is Uniform on [0,1]

If $X$ is uniform on $[0,1]$ and $Y=1/X$ , the reciprocal transformation sends values of $X$ near 1 to values of $Y$ near 1, and values of $X$ near 0 to large values of $Y$ . The density is therefore supported on $[1,\infty)$ and decreases as $y$ grows.

The density is

f_Y(y)=\begin{cases} \dfrac{1}{y^2}, & y\ge 1,\\ 0, & y<1. \end{cases}

Find the support

A uniform random variable on $[0,1]$ satisfies $0<X\le 1$ with probability 1. Taking reciprocals gives $Y=1/X\ge 1$ . There are no values below 1, while arbitrarily large values are possible when $X$ is close to 0.

Strictly speaking, $1/X$ is undefined at $X=0$ . But a continuous uniform random variable equals any particular point with probability zero, including 0. You can define $Y$ to be any value at $X=0$ without changing its probability distribution.

Derive the density from the CDF

For $y<1$ , the event $Y\le y$ is impossible, so $F_Y(y)=0$ . For $y\ge 1$ , use the fact that $X$ is positive almost surely:

\begin{aligned} F_Y(y) &=\Pr(Y\le y)\\ &=\Pr\left(\frac{1}{X}\le y\right)\\ &=\Pr\left(X\ge \frac{1}{y}\right)\\ &=1-\frac{1}{y}. \end{aligned}

The last equality uses the uniform distribution: for $0\le a\le1$ , $\Pr(X\ge a)=1-a$ . Thus the full CDF is

F_Y(y)=\begin{cases} 0, & y<1,\\ 1-\dfrac{1}{y}, & y\ge1. \end{cases}

Differentiating wherever the CDF changes gives

f_Y(y)=F_Y'(y)=\frac{1}{y^2},\qquad y>1.

At the boundary $y=1$ , the density can be assigned the value $1$ according to the displayed formula; changing a density at a single point does not change any probabilities.

Check with the change-of-variables formula

The inverse transformation is $x=1/y$ , and its derivative is $dx/dy=-1/y^2$ . Since $f_X(x)=1$ on $0\le x\le1$ , the one-to-one change-of-variables formula gives

f_Y(y)=f_X(1/y)\left|\frac{d}{dy}\frac{1}{y}\right|=1\cdot\frac{1}{y^2},\qquad y\ge1.

The absolute value matters: a density cannot be negative, even though the inverse function has a negative derivative. This method agrees with the CDF derivation.

Worked probability example

To find the probability that $Y$ lies between 2 and 4, integrate the density:

\begin{aligned} \Pr(2<Y<4) &=\int_2^4\frac{1}{y^2}\,dy\\ &=\left[-\frac{1}{y}\right]_2^4\\ &=\frac{1}{2}-\frac{1}{4}\\ &=\frac{1}{4}. \end{aligned}

You can check this directly using $Y=1/X$ : the event $2<Y<4$ is equivalent to $1/4<X<1/2$ . A uniform variable on an interval of length 1 falls in that interval with probability $1/2-1/4=1/4$ .

The CDF also makes tail probabilities quick to calculate. For $a\ge1$ ,

\Pr(Y>a)=1-F_Y(a)=\frac{1}{a}.

For example, $\Pr(Y>10)=0.1$ . The tail decays slowly enough that the mean is infinite:

\mathbb{E}[Y]=\int_1^\infty y\frac{1}{y^2}\,dy =\int_1^\infty\frac{1}{y}\,dy=\infty.

Common mistakes

Using the wrong support: $Y$ cannot be less than 1, because $X\le1$ . The reciprocal becomes large, not small, near $X=0$ .
Forgetting the Jacobian: Substituting $x=1/y$ into the original density alone is not enough. The derivative factor $|dx/dy|=1/y^2$ is required.
Reversing an inequality incorrectly: From $1/X\le y$ , with $X>0$ and $y>0$ , you get $X\ge1/y$ . The direction changes when taking reciprocals.
Treating $X=0$ as a problem for the distribution: The expression is undefined at exactly 0, but that event has probability zero. It does not add a point mass to the distribution of $Y$ .
Confusing the density with a probability: $f_Y(y)$ is not the probability that $Y$ equals exactly $y$ . For a continuous random variable, that probability is zero; probabilities come from integrating the density over intervals.

References

A First Course in Probability, Sheldon Ross

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