# Seyhmus Güngören

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bio website location Germany age member for 1 year, 7 months seen 7 hours ago profile views 603

$1$ : It means that "there exists something"

$0$ : It doesnt meant that "there exists nothing", instead it means that "there exist something but that is nothing"

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 1d revised How to detemine continuity and differentiablity? added 5 characters in body Dec11 comment Can a density function in a closed ball have an unbounded expected value? Dear Did, I have one more "little" question. When $f$ is standard Gaussian is it also possible to find a density in the closed ball with unbounded expectation? If it is bounded, then I would also wonder some (nice or minimal) conditions on $f$ such that expectation is bounded. should I open a new question for that? I used Lagrangian multipliers for this case and found another Gaussian with a different mean as the density which has the greatest expectation, compared to any other density in the closed ball. Dec10 accepted Can a density function in a closed ball have an unbounded expected value? Dec10 comment Can a density function in a closed ball have an unbounded expected value? okaky understood. Once again thank you very much. Dec10 comment Can a density function in a closed ball have an unbounded expected value? Thank you very much for the answer and the nice example. would there be any example when $\Omega=\mathbb{R}$? This question comes from my comparison of some distance measures. When it is total variation $\int|P-Q|\mathrm{d}\mu$, assume $f$ is standard Gaussian. When I take $g$ as $0.9*f+0.1\mathcal{N}(a,1)$ and when $a\rightarrow\infty$, then $g$ is still in the closed ball of total variation but the expectation also goes to infinity. However, this is not true for the KL divergence. Dec10 revised Can a density function in a closed ball have an unbounded expected value? edited tags Dec10 revised Can a density function in a closed ball have an unbounded expected value? added 2 characters in body Dec10 revised Can a density function in a closed ball have an unbounded expected value? deleted 53 characters in body Dec10 revised Can a density function in a closed ball have an unbounded expected value? added 122 characters in body Dec10 comment Can a density function in a closed ball have an unbounded expected value? $f$ is given and it can be any $f$ with bounded expected value. Yes your understanding is completely correct. For the distance |g(y)-f(y)| I can easily find a counter example but that counterexample doesnt hold for the closed ball with the KL divergence. Dec10 asked Can a density function in a closed ball have an unbounded expected value? Dec6 revised $\int_{t=-\infty}^x (G(t)-F(t))\mbox{d}t\geq 0\forall x$ and $\frac{\mbox{d}F(t)}{\mbox{d}G(t)}$ increasing $\Longrightarrow G(x)\geq F(x)\forall x$? deleted 1 characters in body Nov29 comment is symmetric chi-squared distance “A” metric? @zyx okay thanks again. Nov29 comment is symmetric chi-squared distance “A” metric? @zyx so I need to take the square root of the whole thing I thing it is the same with squared Hellinger distance. We also need to take the square root and it is also symmetric. Thanks for the comment. If you could type an answer I could also accept. Nov29 asked is symmetric chi-squared distance “A” metric? Nov29 comment Power series to calculate LambertW up to infinity? Difficult to understand the question Nov29 revised Could you please help me understand the discrepancy metric? deleted 2 characters in body Nov29 revised Convexity of $\int f(\frac{dP}{dQ})dQ$ for some convex $f$ added 8 characters in body Nov29 revised $\int_{t=-\infty}^x (G(t)-F(t))\mbox{d}t\geq 0\forall x$ and $\frac{\mbox{d}F(t)}{\mbox{d}G(t)}$ increasing $\Longrightarrow G(x)\geq F(x)\forall x$? added 1 characters in body Nov28 revised $\int_{t=-\infty}^x (G(t)-F(t))\mbox{d}t\geq 0\forall x$ and $\frac{\mbox{d}F(t)}{\mbox{d}G(t)}$ increasing $\Longrightarrow G(x)\geq F(x)\forall x$? added 3 characters in body