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Find the value of k for which the given function is a probability density function. f(x = kekx on [0, 2]

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3 votes

Explanation:

I can't understand this question

User Sohilv
by
7.7k points
2 votes
Reading this as


f_X(x)=\begin{cases}ke^(kx)&\text{for }x\in[0,2]\\0&\text{otherwise}\end{cases}

For
f_X(x) to be a valid PDF, the integral over its support must equal 1:


\displaystyle\int_(-\infty)^\infty f_X(x)\,\mathrm dx=\int_(x=0)^(x=2) ke^(kx)\,\mathrm dx=1

Let
y=kx, so that
\frac{\mathrm dy}k=\mathrm dx and the integral becomes


\displaystyle\int_(y=0)^(y=2k)ke^y\,\frac{\mathrm dy}k=\int_0^(2k)e^y\,\mathrm dy=e^y\bigg|_(y=0)^(y=2k)=1

e^(2k)-e^0=1

e^(2k)-1=1

e^(2k)=2

\ln e^(2k)=\ln2

2k=\ln 2

k=\frac{\ln 2}2=\ln\sqrt2
User Mushahid Gillani
by
8.3k points

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