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13 votes
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Suppose $X$ is distributed as Poisson with mean $λ.$ Then $E(1/(X + 1))$ is

  1. $\frac{e^{\lambda }-1}{\lambda }$
  2. $\frac{e^{\lambda }-1}{\lambda +1}$
  3. $\frac{1-e^{-\lambda }}{\lambda}$
  4. $\frac{1-e^{-\lambda }}{\lambda + 1}$
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2 Answers

19 votes
19 votes
Best answer

Answer is (C)

\begin{align*}
E\left( \frac{1}{1+x}\right ) &= \sum_{k = 0}^{\infty} \left( \frac{1}{1+k}\right ) * \frac{\lambda^k*e^{-\lambda}}{k!} \\
 &= \frac{1}{\lambda} * \sum_{k = 0}^{\infty}  \frac{\lambda^{k+1}*e^{-\lambda}}{(k+1)!}\\
 &= \frac{e^{-\lambda}}{\lambda} * \sum_{k = 1}^{\infty}  \frac{\lambda^{k}}{k!}\\
 &= \frac{e^{-\lambda}}{\lambda} * \left( \sum_{k = 0}^{\infty}  \frac{\lambda^{k}}{k!}-1\right)\\
 &= \frac{e^{-\lambda}}{\lambda} * \left( e^{\lambda} -1\right)\\
 &= \frac{1-e^{-\lambda}}{\lambda}

\end{align*}

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4 Comments

so,@ l, for any random variable $Y$ =$f(X)$ ,we can take PMF of $X$ i.e. $P(x)$ for $P(Y)$?

 

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@reena_kandari

No. Not always. If you want to find expected value of Y(where Y is function of X), you can use LOTUS, which does not require you to know PMF of Y. You are still taking sum of elements from X and not Y.

Edit: In my comments I have said PMF. But it equally holds for PDF(i.e. when X is continuous random variable).

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edited by

@Dhruv Patel  yeah :)

@reena_kandari

Suppose, we have a random variable $X$ which can take value as $\{-1,0,1\}$ with $P(X=-1)=0.2$, $P(X=0)=0.5$ and $P(X=1)=0.3$

Then $E[X^2] = 1*P(X=-1)+ 0*P(X=0) + 1*P(X=1) = 1*0.2 + 0*0.5 + 1*0.3$

Now, consider $X^2=Y$, So, random variable $Y$ can take values $\{0,1\}$.

Now, according to you, $E[X^2] = E[Y] = 0*P(Y=0) + 1*P(Y=1)$

Here $P(Y=0) = P(X=0) = 0.5 $ and  $P(Y=1) = P(X= -1) + P(X=1)= 0.5$

Now, put values, you will get the same answer. In 2nd case I have changed the function(Random Variable) but we have to consider probabilities according to our new random variable.

Please tell me if you are not getting it. I will explain through sample space and mapping, maybe from that you can visualize how things are happening here.

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4 votes
4 votes

 

Simplest Solution: Ans(C)

 

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2 Comments

Nice Explanation.

Make correction in the LHS expression!
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First line have error
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