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Question

Explain what is wrong with the statement. The function P(x)=x2exP(x)=x^{2} e^{x} is a cumulative distribution function.

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Answered 6 months ago
Answered 6 months ago
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Consider the wrong statement that P(x)=x2exP(x)=x^{2} e^{x} is a cumulative distribution function.

Now consider the function P(x)=x2exP(x)=x^{2} e^{x} .

Any function is density function if limxP(x)=1,limxP(x)=0,\lim _{x \rightarrow \infty} P(x)=1, \lim _{x \rightarrow-\infty} P(x)=0, and now check the same with the above function,

limxP(x)=limxx2ex\lim _{x \rightarrow \infty} P(x)=\lim _{x \rightarrow \infty} x^{2} e^{x}

Use L-Hospitals' rule here and the above limit will now become,

limxx2ex=limxex(2x+x2)=ex(2()+()2)=\begin{align*} \lim _{x \rightarrow \infty} x^{2} e^{x} &=\lim _{x \rightarrow \infty} e^{x}\left(2 x+x^{2}\right) \\ &=e^{x}\left(2(\infty)+(\infty)^{2}\right) \\ &=\infty \end{align*}

Since, the very first condition is not satisfied, it is not necessary to go for the second one therefore the function P(x)=x2exP(x)=x^{2} e^{x} is not a cumulative distribution function which is stated wrong in the statement that it is a cumulative distribution function.

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