Let [katex] \{ X_n \}_{n\geq1} [/katex] be a sequence of independent and identically distributed random variables having U(0,1) distribution. Let [katex] Y_n = n min\{ X_1,X_2,\dots X_n \} , n \geq 1[/katex]. If [katex] Y_n [/katex] converges to Y in distribution, then the median of Y equals __________________. (round off to 2 decimal places)
The pdf of [katex] Y_n [/katex] is given as:
F_{Y_n}(y) = Pr[ Y_n \leq y] \\
= Pr [ n \quad min\{ X_1,X_2,\dots X_n \} \leq y ] \\
= 1- Pr \left[min\{ X_1,X_2,\dots X_n \} > \frac{y}{n} \right] \\
= 1 - \left( Pr \right[X_1 > \frac{y}{n} \left] \right)^n \\
= 1 - \left( 1 - \frac{y}{n} \right)^n \quad \forall \quad 0 < y < n Now,
1-F_Y(y)={lim}_{n \rightarrow \infty} {1- F_{Y_n}(y)} = {lim}_{n \rightarrow \infty} \left( 1 - \frac{y}{n} \right)^n \\
= e^{-y} \quad \forall \quad 0 < y < \infty \\
\implies F_Y(y) = 1 - e^{-y} \quad 0 < y < \inftyTo find the median we write,
1 - e^{-y} = 0.5 \\
\implies -y = log(0.5) \\
\implies y = 0.3010
Discovering Statistics 
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