GATE 2024 DA Q.11: solution and answer

Official answer: ASource: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. Statement (i): let $X\sim \text{Poisson}(\lambda )$ with $P(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,2,\dots$. Then $E[X]={\sum }_{k\ge 1}k{e}^{-\lambda }{\lambda }^k/k!=\lambda e^{-\lambda }{\sum }_{k\ge 1}{\lambda }^{k-1}/(k-1)!=\lambda$.

2. The factorial moment is $E[X(X-1)]={\sum }_{k\ge 2}{e}^{-\lambda }{\lambda }^k/(k-2)!={\lambda }^2{e}^{-\lambda }{\sum }_{j\ge 0}{\lambda }^j/j!={\lambda }^2$.

3. So $E[X^2]=E[X(X-1)]+E[X]={\lambda }^2+\lambda$.

4. Then $\mathrm{Var}(X)=E[X^2]-(E[X]{)}^2={\lambda }^2+\lambda -{\lambda }^2=\lambda$, which equals the mean. Statement (i) is true.

5. Statement (ii): the normal distribution $N(\mu ,{\sigma }^2)$ has mean $\mu$ and variance ${\sigma }^2$. The standard normal is the case $\mu =0$, ${\sigma }^2=1$, so its mean is zero and its variance is one. Statement (ii) is true.

6. Both statements are true, so the answer is option (A).