Update 2.md

Author: koohong

_problems/unit-02/E-cdf-of-a-continuous-rv/2.md

@@ -5,4 +5,4 @@ statement: |
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 Sol:
-Given CDF, we can find its pdf by $\frac{d}{dx}[e^{\frac{-1}{x}}]=\frac{e^{frac{-1}{x}{x^2}}$ and $\int_{0}^{\infty}\frac{e^{frac{-1}{x}{x^2}}dx=1$
+Given CDF, we can find its pdf by $\frac{d}{dx}[e^{\frac{-1}{x}}]=\frac{e^{frac{-1}{x}}{x^2}$ and $\int_{0}^{\infty}\frac{e^{frac{-1}{x}}{x^2}dx=1$