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specified that the answer holds for a particular case.
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edit approved Jan 3, 2014 at 13:51
splinter123
edit approved Jan 3, 2014 at 13:51
splinter123
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YesThe answer is yes for $d=1$. By denoting $F$ the cdf of $X^1$, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ \left|\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP\right|\le \left|\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP\right|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.

Yes, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ \left|\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP\right|\le \left|\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP\right|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.

The answer is yes for $d=1$. By denoting $F$ the cdf of $X^1$, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ \left|\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP\right|\le \left|\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP\right|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.

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edit approved Jan 2, 2014 at 21:11
András Bátkai
edit approved Jan 2, 2014 at 21:11
András Bátkai
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Yes, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theoremGlivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ |\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP|\le |\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP|+|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP|+|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP|$$$$ \left|\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP\right|\le \left|\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP\right|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.

Yes, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ |\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP|\le |\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP|+|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP|+|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.

Yes, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ \left|\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP\right|\le \left|\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP\right|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.

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adrido
adrido
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Yes, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$ |\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP|\le |\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP|+|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP|+|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP|$$

Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.