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The probability density function of a random variable X is symmetric about 0. Then

$$\int_{-2}^2 \int_{-\infty}^x f(u) du dx$$ is?

I figured that integral of a PDF from $-\infty$ to $x$ would be it's cdf. But how should I integrate next?

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    Should it be $f(u,x)$? – 2017-02-08
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    No it's just $f(u)$ in the question. – 2017-02-08
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    A vocabulary issue: You should say that the pdf is an **even** **function**, and keep "symmetric" (with respect to $y$ axis) for qualifying its **curve**. – 2017-02-08
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    Another little grammatical remark: in your title, you should write "its" instead of "it's", which is an abbreviation "it is". – 2017-02-08

2 Answers 2

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Hints:

  • Let $F(x) = \int_{-\infty}^x f(u) du$. Then, we have \begin{align*} F(x) &= \int_{-\infty}^x f(u) du \\ &= \int_{-\infty}^x f(-u)(-1)(-1) du \\ &= - \int_{\infty}^{-x} f(u) du \\ &= \int_{-x}^{\infty} f(u) du \\ &= 1 - F(-x). \end{align*} That is, the function $$ G(x) = F(x) - \frac12 $$ is odd around $0$.

  • Now, what is $$ \int_{-2}^2 F(x) dx = \int_{-2}^2 \frac12 + G(x) dx? $$

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    [+1] fine hint. – 2017-02-08
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{-2}^{2}\int_{-\infty}^{x}\mrm{f}\pars{u}\,\dd u\,\dd x & = \int_{-2}^{2}\int_{-\infty}^{2}\bracks{u < x}\mrm{f}\pars{u}\,\dd u\,\dd x = \int_{-\infty}^{2}\mrm{f}\pars{u}\int_{-2}^{2}\bracks{x > u}\,\dd x\,\dd u \\[5mm] & = \int_{-\infty}^{2}\mrm{f}\pars{u} \braces{\bracks{u < 0}\int_{-2}^{2}\,\dd x + \bracks{0 < u < 2}\int_{u}^{2}\,\dd x}\,\dd u \\[5mm] & = 4\int_{-\infty}^{0}\mrm{f}\pars{u} + \int_{0}^{2}\mrm{f}\pars{u}\pars{2 - u}\,\dd u \end{align}