Let X,Y,Z be independent and N(1,1). Find the probability that X+Y≥3Z using convolution.
I did the following: Let W=X+Y−3Z,
then we have that
W∼N(1+1−3,1+1+9)=N(−1,11).
so
P(X+Y≥3Z)=P(W≥0)=1−P(W≤0)=1−FW(0)=1−Φ(
1
√
11
)≈≈1−0.618=0.382.
However, I'm in the chapter treating convolution. How can I solve this problem using this approach instead?
then we have that
W∼N(1+1−3,1+1+9)=N(−1,11).
so
P(X+Y≥3Z)=P(W≥0)=1−P(W≤0)=1−FW(0)=1−Φ(
1
√
11
)≈≈1−0.618=0.382.
However, I'm in the chapter treating convolution. How can I solve this problem using this approach instead?
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