jakemf1986
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This is not actually a homework question but something I saw in a (economics) journal article and have been having trouble getting. It's very simple so I thought I'd post it here.
Define the variable V (implicitly) by:
[itex]V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s)[/itex],
where r is a constant and F has support on [0,∞).
Question: Show that [itex]\frac{dV}{db}=\frac{1}{1-rF(rV)}[/itex],
My attempted solution: Differentiate with respect to V and use Leibniz's Rule to get
[itex]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)[/itex]
Rearrangement yields
[itex]\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}[/itex]
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:
[itex]V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s)[/itex],
where r is a constant and F has support on [0,∞).
Question: Show that [itex]\frac{dV}{db}=\frac{1}{1-rF(rV)}[/itex],
My attempted solution: Differentiate with respect to V and use Leibniz's Rule to get
[itex]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)[/itex]
Rearrangement yields
[itex]\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}[/itex]
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.