Question

If X is uniform on [10,20], find the approximate and exact mean and variance of Y=1/X, and compare them.

Answer 1

Finding the distribution of \(Y\):
Use the definition of the cdf, and then differentiate to find the pmf.
\[P(Y\le y)=P(1/X\le y)\\
=P(X\ge1/y)
\\ =1-P(X\le1/y)\\
=1-F_X(1/y)\]
Differentiate both sides w.r.t. y to find the pdf of \(Y\)
\[\frac{d}{dy}\underbrace{P(Y\le y)}_{F_Y(y)}=\frac{d}{dy}\left((1-F_X(1/y)\right)
\\
f_Y(y)=-f_X(1/y)\frac{-1}{y^2}=\frac{1}{y^2}f_X(1/y)\]
So, now the pdf of X, \(f_X(x)\) is \(\large \frac{1}{20-10}=\frac{1}{10} \)
So, \[f_Y(y)=\frac{1}{y^2} \frac{1}{10}\] you don't have to worry about the 1/y in f(x) in the argument since we have a uniform distribution, so the pdf is the same for all y.

Now with this, you just apply the definition of the expectation and variance and do the appropriate integrals.
\[E(Y)=\int_{10}^{20}y\cdot\frac{1}{y^2} \frac{1}{10}dy\]
\[E(Y^2)=\int_{10}^{20}y^2\cdot\frac{1}{y^2} \frac{1}{10}dy\]
And \[Var(Y)=E(Y^2)-E(Y)\]