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$E[X^2|X>1] = E[(X+1)^2]$

By the memoryless property, the conditional distribution of $X$, given that $X>1$, is the same as the unconditional distribution of $X+1$. Therefore, the equation stands.

I'm trying to prove $$P(X|X>1) = P(X+1)$$ as given in the statement. However, the equation doesn't come as true according to my calculation. Please let me know where my mistake is. Here is my calculation:

$\begin{align}P(X|X>1) &= \frac{P(X=k)}{P(X>1)} , \quad k>1\\ &= \frac{\lambda e^{-\lambda k}}{e^{-\lambda}}\\ &= \lambda e^{-\lambda k + 1},\quad k>1 \end{align} $

$\begin{align}P(X+1) &= \lambda e^{-\lambda (x+1)}\\ &= \lambda e^{-\lambda x - \lambda } \end{align} $

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underdisplayname
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  • Your tag and title mention it, but you may want to say $X$ is an exponential distribution and state the mean of the distribution. – Lockjaw Jul 03 '19 at 02:01
  • it depends on the distribution of $X$. You are assuming that $X$ is discrete writing $P(X=k|X>1)$. Moreover: what means $P(X|X>1)$? Probably you wanted to write $P(X=k|X>1)$ – Masacroso Jul 03 '19 at 02:02

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What we really want to prove is, for $k\ge1$ (since it is trivial when $k<1$), $$P(X=k|X>1) = P(X+1=k)$$ which expands into $$\frac{\lambda e^{-\lambda k}}{e^{-\lambda}}=\lambda e^{-\lambda(k-1)}$$ Simplifying the LHS leads to the RHS, proving the identity: $$\frac{\lambda e^{-\lambda k}}{e^{-\lambda}}=\lambda e^{-\lambda k-\lambda}=e^{-\lambda(k-1)}$$

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