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Suppose real-valued random variables $\{X_{n}\} $ converges to $X$ in distribution. Then, will the quantile of the distribution of $\{X_n\}$ converge to the quantile of $X$? .

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Yes. If $X$ is a random variable with distribution function $F$, then for $0

Added: It's a nice exercise to prove this result from the definition. On the other hand, it is Proposition 5 (page 250) in A Modern Approach to Probability Theory by Bert Fristedt and Lawrence Gray, and is also proved in Chapter 21 of Asymptotic Statistics by A. W. van der Vaart.

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    But what if $Q$ is not continuous at $p$?2011-12-25
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    Suppose we know that $F$ is strictly increasing. Will this additional condition help?2011-12-25
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    The discontinuity points of $Q$ correspond to the "flat spots" on the graph of $F$ (Draw a picture!). So if $F$ is strictly increasing, then $Q$ is continuous at all $02011-12-25
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    Yes. But what will be the formal proof?2011-12-25
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    By the way, if $F$ is strictly increasing on the support of $X$, and the supports of all $X_n$ are contained in the support of $X$, then $Q$ is continuous on $[0,1]$, and the convergence of the quantile functions is uniform. This simple observation has applications in the theory of Toeplitz matrices. http://dx.doi.org/10.1007/s40590-016-0105-y2016-10-23