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Given that $U_1,U_2$ are $iid$ Uniform(0,1), find the variance of $$T=\frac{\ln{U_1}}{\ln U_1 + \ln(1-U_2)}$$

Let $U\sim Uniform(0,1)=U(0,1)$

Multiply numerator and denominator by $-1$. Since $1-U\sim U(0,1)$, let $1-U_2=U_3 \sim U(0,1)$: $$T=\frac{-\ln{U_1}}{-\ln U_1 - \ln(U_3)}$$

Convert to exponential, and write the exponential as a gamma: $$T=\frac{X}{X+Y}, X, Y \sim exp(1)=Gamma(1,1)$$

Use the property that if X and Y are independent such that: $X \sim Gamma(\alpha, θ), Y \sim Gamma(β, θ)$ then $ X+Y∼Beta(α,β)$ $$T \sim Beta(1,1)$$

Since $Beta(1,1) \sim U(0,1)$ $$V(T) = \frac{(b-a)^2}{12}=\frac{(1-0)^2}{12}=\frac{1}{12}$$

Since the final distribution is standard uniform, I wanted to know if there is an easier solution that is not so circuitous

Gabriel Romon
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Starlight
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1 Answers1

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It suffices to compute the CDF of $T$, which has the same distribution as $$T' =\frac{\ln{U_1}}{\ln U_1 + \ln U_2}$$ Clearly $T'$ has values in $(0,1)$ a.s. and if $t\in (0,1)$: $$\mathbb P(T'\leq t) = \int_0^1\int_0^1 1_{\frac{\ln x}{\ln x +\ln y}\leq t} \text{dy dx} $$

Since $\frac{\ln x}{\ln x + \ln y}\leq t \iff y\leq x^{\frac{1}{t} - 1}$, $$\mathbb P(T'\leq t)=\int_0^1x^{\frac{1}{t} - 1}dx = \frac {1^{{\frac{1}{t} - 1+1}}}{\frac{1}{t} - 1+1}=t.$$

Therefore $T$ has the same CDF as $\mathcal U(0,1)$, and $T\sim \mathcal U(0,1)$.

Gabriel Romon
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  • I understand double integration, but I didn't really understand how you integrated with respect to y and then x. What are the limits, and how do you get them? – Starlight Jan 09 '25 at 11:07
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    @Starlight Prove$$\frac{\ln x}{\ln x+\ln y}\le t\iff y\le x^{(1-t)/t}$$(take care to note $\ln x,,\ln y\le0$). – J.G. Jan 09 '25 at 11:35
  • @J.G. It took a few steps, but I proved it. – Starlight Jan 09 '25 at 12:50
  • If $y\leq x^{(1-t)/t}$, and $t\in(0,1)$, then $$t\rightarrow0 \implies x^{(1-t)/t}\rightarrow\infty$$, and that goes outside the region of integration (0<y<1). How do we address that? – Starlight Jan 10 '25 at 04:21
  • @Starlight What do you mean ? $y$ is in the inner integral, hence it is integrated out and $\int_0^1 1_{\frac{\ln x}{\ln x +\ln y}\leq t} dy = x^{(1-t)/t}$. Next, you integrate over $x$: $\int_0^1x^{\frac{1}{t} - 1}dx$. – Gabriel Romon Jan 11 '25 at 16:20
  • What I meant was that in the inner integral we get $y\leqx^((1-t))/t$, and that seems to go outside the region of integration (which is 0<y<1) when $t$ is very close to zero, and hence $y<1/x$. – Starlight Jan 11 '25 at 17:49
  • @Starlight Since $t\in (0,1)$ and $x\in (0,1)$, we have $\frac{1-t}t \ln x\leq 0$ and thus $0\leq x^{(1-t)/t}\leq 1$. There is no cause for concern. – Gabriel Romon Jan 11 '25 at 20:50
  • Got it. Thanks for the detailed responses. – Starlight Jan 12 '25 at 03:35