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I'm attempting to differentiate the following double integral with respect to $u$:

$$I(u) = \int_a^u\int_b^v [(y-u) + (v - x)] f(x,y)\,dx \,dy$$

where $f(x,y)$ is the joint density function of RV $X$ and $Y$.

I'm trying to find $I'(u)$, that is, the derivative of $I$ with respect to $u$. I know this involves Leibniz integral rule, however I'm getting stuck on the double integral part.

If I let $g(x,y,u) = \int_b^v [(y-u) + (v - x)] f(x,y)\,dx$ then I think the problem to solve is:

$$\frac{d}{du}\bigg(\int_a^ug(u,x,y)\,dy\bigg) = g(u,x,u) + \int_a^u\frac{\partial}{\partial u}g(u,x,y)\,dy$$

However I'm getting stuck evaluating the two expressions on the RHS.

Any help?

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Your approach is correct, so I'm going to let $g(u,v,y) = \int_b^v [(y-u)+(v-x)]f(x,y)dx$ (note that this is not actually a function of $x$ since it gets integrated over, I think you may have had a typo). Then by the Liebnitz rule we have: $$ \frac{d}{du} \int_a^u g(u,v,y)dy = g(u,v,u) + \int_a^u \frac{d}{du} g(u,v,y) dy $$ The first part simplifies nicely: $$ g(u,v,u) = \int_b^v [(y-u) + (v-x)]f(x,y)dx = \int_b^v (v-x)f(x,y) dx $$ and the partial derivative of $g$: $$ \frac{d}{du} g(u,v,y) = \frac{d}{du}\int_b^v [(y-u) + (v-x)]f(x,y)dx = \int_b^v \frac{d}{du}[(y-u) + (v-x)]f(x,y)dx = -\int_b^v f(x,y) dx $$

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  • $\begingroup$ I made a typo in the OP. The $f(x,y)$ should be distributed through to $(y-u)$. I think this just changes the last line of your answer a bit. $\endgroup$ – measure_theory Jul 31 '18 at 17:32
  • $\begingroup$ OK I shall edit accordingly. $\endgroup$ – Dark Malthorp Jul 31 '18 at 17:34

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