Using law of iterated expectations

Question

. I think have the Poisson and uniform distributed equation which would be $e^{-10} (10^n/n!)$ and $1/50$ but I'm not sure how to apply the law and help would be appreciated.

Answer 1

Define $Y=\sum_{i=1}^NX_i$, hence $$E\{Y\} {= E\left\{\sum_{i=1}^NX_i\right\} \\= E_N\left\{E\left\{\sum_{i=1}^NX_i\Bigg|N\right\}\right\} \\= \sum_{n=50}^{100}\Pr\{N=n\}E\left\{\sum_{i=1}^NX_i\Bigg|N=n\right\} \\= \sum_{n=50}^{100}{1\over 51}\cdot E\left\{\sum_{i=1}^nX_i\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} E\left\{\sum_{i=1}^nX_i\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} 10n \\=750\$ } $$ where we have used the fact that the summation of two Poisson r.v.'s with parameters $\lambda_1$ and $\lambda_2$ is a Poisson r.v. with parameter $\lambda_1+\lambda_2$.

The variance is similarly calculated as $$ E\{Y^2\} {= {1\over 51}\cdot\sum_{n=50}^{100} E\left\{\left[\sum_{i=1}^nX_i\right]^2\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} E\left\{\sum_{i,j}X_iX_j\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} E\left\{\sum_{i=j}X_iX_j\Bigg|N=n\right\} \\+ {1\over 51}\cdot\sum_{n=50}^{100} E\left\{\sum_{i\ne j}X_iX_j\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} E\left\{\sum_{i=1}^nX_i^2\Bigg|N=n\right\} \\+ {1\over 51}\cdot\sum_{n=50}^{100} \sum_{i\ne j}E\left\{X_iX_j\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} 10n \\+ {1\over 51}\cdot\sum_{n=50}^{100} \sum_{i\ne j}E\left\{X_i\Bigg|N=n\right\} E\left\{X_j\Bigg|N=n\right\} \\= {1\over 51}\cdot\sum_{n=50}^{100} 10n \\+ {1\over 51}\cdot\sum_{n=50}^{100} \sum_{i\ne j}100 \\= E\{Y\}+ {1\over 51}\cdot\sum_{n=50}^{100} 100(n^2-n) \\=577416{2\over 3}\$^2 } $$ hence $${\sigma_Y^2=14916{2\over 3}\$^2\\\sigma_Y\approx 122.13\$}$$