# Uniform(0,$\theta$) ratio UMVUE?

Let $$X_i$$ be i.i.d $$uniform(0,\theta)$$ and $$Y_i$$ be i.i.d $$uniform(0,\lambda)$$.

The problem is to find the UMVUE of $$\frac{\theta}{\lambda}$$.

My attempt has been to use the fact that $$(X_{(n)},Y_{(n)})$$, the $$n^{th}$$ order statistics, are jointly sufficient and complete. Thus any unbiased estimator based on them must be UMVUE. Unfortunately, as far as I can tell, no unbiased estimator of $$\frac{1}{\lambda}$$ exists and so the $$X_i$$'s and $$Y_i$$'s cannot be treated separately.

It is easy to verify that the maximums of these unform samples have pdf's

$$f_X(x) = nx^{n-1}/\theta^n \quad \quad f_Y(y) = ny^{n-1}/\lambda^n$$

Thus if $$\delta(X_{(n)},Y_{(n)})$$ is an unbiased estimator based then it must satisfy

$$\int_0^\theta \int_0^\lambda \frac{\delta (x,y) n^2 x^{n-1}y^{n-1}}{\theta^n \lambda^n} dy dx = \frac{\theta}{\lambda}$$

Or equivalently

$$\int_0^\theta \int_0^\lambda \delta (x,y) x^{n-1}y^{n-1} dy dx = \frac{\theta^{n+1}\lambda^{n-1}}{n^2}$$

Differentiating both sides with respect to $$\theta$$ then gives

$$\int_0^\lambda \delta(\theta,y)\theta^{n-1}y^{n-1}dy = \frac{(n+1)\theta^n \lambda^{n-1}}{n^2}$$

Differentiating once more with respect to $$\lambda$$ yields

$$\delta(\theta,\lambda) \theta^{n-1}\lambda^{n-1} = \frac{(n+1)(n-1)\theta^n\lambda^{n-2}}{n^2}$$

So that

$$\delta (\theta,\lambda) = \frac{(n+1)(n-1)}{n^2}\times \frac{\theta}{\lambda}$$

Hence the UMVUE is given by

$$\delta (X_{(n)},Y_{(n)}) = \frac{(n+1)(n-1)}{n^2}\times \frac{X_{(n)}}{Y_{(n)}}$$

Which seems sensible. Is this working correct?

Your working looks entirely correct to me. We can also confirm this result by an alternative method. Let us posit that the UMVUE is likely to be some scaled version of the ratio statistic:

$$R_n \equiv \frac{X_{(n)}}{Y_{(n)}}.$$

For all $$r > 0$$ we have:

\begin{aligned} F_{R_n}(r) \equiv \mathbb{P}(R_n \leqslant r) &= \mathbb{P}( X_{(n)} \leqslant r Y_{(n)} ) \\[6pt] &= \int \mathbb{P}(X_{(n)} \leqslant r y) f_Y(y) \ dy \\[6pt] &= \int \limits_0^\lambda \min \Big(1, \frac{ry}{\theta} \Big)^n \cdot \frac{n y^{n-1}}{\lambda^n} \ dy \\[6pt] &= \begin{cases} \frac{1}{2} \big( \frac{\lambda r}{\theta} \big)^n & & & \text{if } r \leqslant \tfrac{\theta}{\lambda}, \\[6pt] 1 - \frac{1}{2} \big( \frac{\theta}{\lambda r} \big)^n & & & \text{if } r > \tfrac{\theta}{\lambda}. \\ \end{cases} \end{aligned}

Thus, for $$n>1$$ we have the expectation:

\begin{aligned} \mathbb{E}(R_n) &= \int \limits_0^\infty (1-F_{R_n}(r)) \ dr \\[6pt] &= \int \limits_0^{\theta / \lambda} \Big( 1-\frac{1}{2} \Big( \frac{\lambda r}{\theta} \Big)^n \Big) \ dr + \int \limits_{\theta / \lambda}^\infty \frac{1}{2} \Big( \frac{\theta}{\lambda r} \Big)^n \ dr \\[6pt] &= \frac{\theta}{\lambda} \bigg( 1-\frac{1}{2(n+1)} \bigg) + \frac{\theta}{\lambda} \frac{1}{2(n-1)} \\[6pt] &= \frac{\theta}{\lambda} \bigg( 1-\frac{1}{2(n+1)} + \frac{1}{2(n-1)} \bigg) \\[6pt] &= \frac{\theta}{\lambda} \cdot \frac{2(n-1)(n+1) - (n-1) + (n+1)}{2(n-1)(n+1)} \\[6pt] &= \frac{\theta}{\lambda} \cdot \frac{n^2}{n^2-1}. \\[6pt] \end{aligned}

Thus, for all $$n>1$$ we have the unbiased estimator:

$$\widehat{\theta / \lambda} = \frac{n^2-1}{n^2} \cdot R_n,$$

which is what you have derived in your own working.