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I am taking my first stats course and struggling quite a bit as my professor does not explain things well. I wanted to double check that I am on the right track. My question is: Show that $f(x)≥0$ for all $x ∈ R$ , where $x$ is the exponential density function:$$f(x) = \begin{cases}λ\operatorname e^{−λx} & \text{if }x ≥ 0,\\ 0 & \text{if }x<0\end{cases}$$

Could I prove this by taking the limit of the exponential density function both as x approaches negative and positive infinity?

Martin Sleziak
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    The exponential function is always positive. – Novice Feb 05 '23 at 22:58
  • Your problem comes down to showing $e^t>0$ where $x\in \mathbb{R}$. You can definitely show this using the method you described (with a couple of caveats). You could show that for an arbitrarily large negative $t$ the function is positive, and that it is increasing. – Numeral Feb 05 '23 at 23:10
  • Or.... assume the opposite... that there is at least one $x$ for which $e^{- \lambda x}$ is negative. By continuity, there must then be another (smaller) $x^$ for which $e^{- \lambda x^} = 0$. Take the logarithm of both sides... – David G. Stork Feb 05 '23 at 23:24
  • @Numeral I see what you are saying. For an arbitrarily large t, the function is obviously positive and increasing. Then I am thinking of showing as x approaches positive infinity to prove that it approaches zero from that direction as well. –  Feb 05 '23 at 23:33
  • By definition, the density is always non-negative. – van der Wolf Feb 06 '23 at 10:12

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