- #1
jakemf1986
- 7
- 0
A simple computational question, that I saw in a journal article and have been having trouble getting.
Define the variable V (implicitly) by:
V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s),
where r is a constant and F has support on [0,∞).
Question: Show that \frac{dV}{db}=\frac{1}{1-rF(rV)},
My attempted solution: Differentiate with respect to V and use Leibniz's Rule to get
1=\frac{db}{dV} + r\cdot rVf(V) - 0 + \int^{rV}_0 r dF(s) + 0 - rVf(V) + 0 = \frac{db}{dV} + r^{2} Vf(V) + rF(rV) - rVf(V)
Rearrangement yields
\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}
Notice that my solution has an additional ugly term in the denominator.
Is my solution wrong? Or could perhaps a mistake have been made in the article?
Thank you.
Define the variable V (implicitly) by:
V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s),
where r is a constant and F has support on [0,∞).
Question: Show that \frac{dV}{db}=\frac{1}{1-rF(rV)},
My attempted solution: Differentiate with respect to V and use Leibniz's Rule to get
1=\frac{db}{dV} + r\cdot rVf(V) - 0 + \int^{rV}_0 r dF(s) + 0 - rVf(V) + 0 = \frac{db}{dV} + r^{2} Vf(V) + rF(rV) - rVf(V)
Rearrangement yields
\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}
Notice that my solution has an additional ugly term in the denominator.
Is my solution wrong? Or could perhaps a mistake have been made in the article?
Thank you.
