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Question

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

Solution

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Consider the wrong statement that $P(x)=x^{2} e^{x}$ is a cumulative distribution function.

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

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

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

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

\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 $P(x)=x^{2} e^{x}$ is not a cumulative distribution function which is stated wrong in the statement that it is a cumulative distribution function.

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